Inverse Trigonometric functions | Part 1 | JEE Advanced compendium | Basics, Domain, Range & Graphs

0:00 today we'll be discussing trigonometric inverse functions now to understand trigonometric inverse function first we have to understand trigonometric function so what is the role of a trigonometric function any trigonometric function say y equals sine x it takes any angle x as an input and gives the value of sine as the output suppose we take is zero so when x is zero sine zero is zero if we take angle as pi by six sine x will give us one by two at pi by 4 we'll get the value of sine is

0:46 1 by root 2 at pi by 3 we'll be getting this value as root 3 by 2 and at pi by 2 we'll get the value of sine x is 1. now what about trigonometric inverse function now role of inverse function will be opposite to that of this trigonometric function now when we write trigonometric inverse function we write them as sine inverse of x or we write them as r sine x now here the input will be values of x and output will be the angles so we'll

1:32 put x as 0 we'll get sine inverse x is 0 if we'll put x as 1 by 2 we'll get sine inverse x is pi by 6 and we'll put this value as 1 by root 2 we'll get value of sine inverse x is pi by 4. so any trigonometric inverse function is nothing but an angle so from now on whenever you see any trigonometric inverse function just think of it as an angle it is nothing but some angle theta now one problem that still remains is suppose we wish to find the value of sine inverse 1 by 2. so we need to find an angle at which the value of sine theta is 1 by 2 so we know that value of sine theta is 1 by 2 at pi by 6 but this value is also 1 by 2 at 5 pi by 6 and

2:19 many other different angles now definition of function says for any one valid input there has to be only one output in this case we are getting multiple outputs so still it is not a function so how do we define tegrometric inverse function now in order to be able to define inverse trigonometric function we have to go back to functions suppose we have any function f defined from a to b y equals fx now if this function is 1 1 and on 2 then we say this function is invertible and we define inverse of this function

3:04 as f inverse from b to a y equals f inverse of x and if we have to draw the graph of f inverse x from the graph of f x all we need to do is we need to take a reflection of graph of y equals fx about y equals x line so using this fundamental concept we'll try and define trigonometric inverse functions and we'll also draw the graphs now the first trigonometric inverse function is y equals

3:51 sine inverse of x now for sine inverse x our basic function is sine x now we have to define sine x in the interval so that it is 1 1 and on 2 so what we'll do is we'll define function sine x in the interval minus pi by 2 to plus pi by 2 and its range will be minus 1 and plus 1 and this is how we can write y equals sine x now we'll draw this graph that's the graph of y equals sine x from minus pi by 2 to plus pi by 2

4:38 and this is plus 1 and this is -1 and in this interval this function is both 1 1 and 1 2 now when this function is both 1 1 and on 2 we can now define its inverse function so its inverse function will be defined from b to a so we have to write sine inverse we write f inverse from minus 1 to 1 to minus pi by 2 to plus pi by 2 and we'll write this as y equals sine inverse of x so this is how we fundamentally define inverse trigonometric functions now this

5:24 is domain of sine inverse and this is the range and we call this as principal value range so when we study trigonometric inverse function the first and the foremost important thing is its principal value range so for each of the six trigonometric functions you must know its domain as well as its principal value range we could have defined this function from pi by 2 to 3 pi by 2 then we could have got a different range so for different intervals we will get different range but we have defined these functions such that this range from minus pi by 2 to plus pi by 2 it becomes our standard

6:11 range which we call principal value range now we have to draw the graph of sine inverse of x now one way of drawing this graph can be invert the values on x axis and y axis so here will be minus 1 plus 1 here it is pi by 2n minus pi by 2 now if we look at this point this point is pi by 2 comma 1 now we'll interchange x and y so here it will be 1 comma pi by 2 so i'll mark this point so this point is 1 comma pi by 2 this point is 0 0 so if i interchange x and y this will

6:57 again be 0 comma 0 and this point it is minus pi by 2 minus 1 now it will be minus 1 minus pi by 2 so we'll draw a freehand sketch now this is the graph of sine inverse of x and as it is evident from the graph this sine inverse x it is an increasing function we need this information when we are going to solve inequality problems now the second one is y equals cos inverse of x so for this we will define cos x from 0 to pi and range of cos x is

7:45 minus 1 to 1 and this is y equals cos x so cos 0 is 1 cos pi by 2 is 0 and then cos pi is minus 1 so that is the graph of cos x in the interval 0 to pi now cos inverse it'll be defined from minus 1 to 1 to 0 to pi so principal value range of cos inverses from 0 to pi and we'll write this as y equals cos inverse of x now we have to draw the graph of cos inverse we'll plot these values on x axis and y

8:31 axis so on x axis we'll have minus 1 plus 1 and 0 and on y axis we'll have pi by 2 and pi now this point it is pi comma minus 1 here it will be minus 1 comma pi so at minus 1 this value will be pi now this is pi by 2 comma 0 here will be 0 pi by 2 and this point is 0 comma 1 and i will be 1 comma 0. so we'll draw a freehand sketch that's the graph of y equals cos inverse of x now the third one is

9:18 y equals 10 inverse of x so again we'll define 10 inverse we'll define it in such a way that it will be 1 1 and onto so this is y equals 10 x pi by 2 minus pi by 2 0 now for 10 inverse we'll define it from r to minus pi by 2 to plus pi by 2

10:03 and lb y equals 10 inverse of x now we will draw this graph now previously we have two vertical asymptotes here we'll have two horizontal asymptotes so this is pi by 2 and this is minus pi by 2. this is 0 comma 0 so it'll still be 0 comma 0 now in this case when x is pi by 2 y is infinite so here it will be when x is infinite y is pi by 2 so at infinite it'll be pi by 2 and the same way at minus pi by 2 it was minus infinite so here at minus infinite it

10:51 will be minus pi by 2 so this graph will start from here at 0 it will be 0 and then at infinite it'll be pi by 2 so that is the graph of y equals 10 inverse of x with its principal value range at minus pi by 2 n plus pi by 2. now we come to y equals cot inverse of x now for cot inverse we will define cortex from 0 to pi range is r this is y equals cortex this is 0

11:36 pi by 2 n pi and that's the graph of cortex now we'll define cot inverse and it'll be defined from r to 0 to pi and lb y equals cot inverse of x now we'll mark values on x axis and y axis so this is pi by 2 and we'll have an asymptote at pi now this point is pi by 2 comma 0 here will be 0 pi by 2 at 0 it is plus infinite so at infinite it will now be 0

12:22 and at pi it has minus infinite so at minus infinity it will be pi so this is the graph of y equals cot inverse of x with its principal value range at 0 to pi now fifth one is y equals cosec inverse of x so for this we will define this function from minus pi by 2 to plus pi by 2 except 0 and its range will be from minus infinity to minus 1 union

13:07 1 to infinite now we'll draw this graph so this is minus pi by 2 plus pi by 2 this is plus 1 and this is minus 1. so that is the graph of y equals cosec x in the interval minus pi by 2 to plus pi by 2 except 0 and its ranges from minus infinity to minus 1

13:53 union 1 to infinite we'll define this function f inverse from minus infinite to minus 1 union 1 to infinite to minus pi by 2 to plus pi by 2 except 0 and it'll be y equals cosec inverse of x now we'll have to draw this graph so this is 0 we'll have asymptote set pi by 2 n minus pi by 2

14:40 and then we'll also mark 1 and plus 1 now this point is pi by 2 comma 1 here will be 1 comma pi by 2 at 1 value is pi by 2 now this is when x is 0 y is infinite so it infinite this value will be 0 so we'll join this this point is minus pi by 2 minus 1 now here will be minus 1 minus pi by 2. so at minus 1 this value will be minus pi by 2 and here when x is 0 y is minus 1 8 so at minus infinite this value will be 0. so that is the graph of y equals cosecant inverse of x and then finally we'll have our sixth graph which is

15:27 y equals secant inverse of x and for that we will define secant theta in the interval zero to pi except at pi by 2 and its range will be from minus infinite to minus 1 union 1 to infinite so we'll draw this graph so this is plus 1 and minus 1 this is pi by 2 and this is pi now graph of secant x is

16:16 this one so this is your function y equals secant x now we'll define secant inverse and it'll be defined from minus infinity to minus 1 union 1 to infinite to 0 to pi minus pi by 2 and it'll be y equals secant inverse of x and now we'll draw its graph so we'll mark the points so 0 pi by 2 pi asymptote at pi by 2 and we have minus 1 and

17:02 plus 1. now this point is 0 comma 1 so here will be 1 comma 0 so 1 this value will be 0 now when x is pi by 2 y is infinite so when x is infinite now y will be pi by 2 so this way now this point is pi comma minus 1 so at minus 1 lb pi and when x is pi by 2 y is minus infinity so at minus infinity it will be pi by 2. so that is the graph of secant inverse of x and its principal value ranges from 0 to pi minus pi by 2. so this is how we

17:48 define and draw the graphs of trigonometric inverse functions for all the six trigonometric functions you must know their domain and most importantly their principal value range so for sine inverse x principle value ranges minus pi by 2 to plus pi by 2 which makes fourth and first quadrant for cos inverse principal value ranges from zero to pi which is first and second quadrant for 10 inverse it is again fourth and first from minus pi by 2 to plus pi by 2 both not inclusive for cot inverse it is again first and second for cosec inverse x it is again

18:34 fourth and first again from minus pi by 2 to plus pi by 2 except at 0 and for secant inverse it is 0 to pi except pi by 2 which again is first and second quadrant so whenever we are going to write any trigonometric inverse function we think of this trigonometric inverse function as some angle theta and we need to ensure that this value of theta it must lie in principle value range unless otherwise stated in the problem so now when we write sine inverse 1 by 2 so we know that this angle it must lie between minus pi by 2 and plus pi by 2 so in this interval the only possible solution is

19:21 pi by 6 in the same way if we have to write sine inverse minus 1 by 2 now in the fourth quadrant value of sine theta will be minus 1 by 2 at minus pi by 6 so this is how we are going to use this principal value range now there are questions based on principal value range which at first appear very difficult but then the simple problem say for example if sine inverse x plus sine inverse y plus sine inverse z equals 3 pi by 2 then find the value of

20:07 x to the power 2 0 to 0 plus y to the power 2 0 to 0 z to the power 2 0 2 0 plus 1 upon x to the power 2 0 2 0 plus 1 upon y to the power 2 0 2 0 plus 1 upon z to the power 2 0 2 0 now we have not studied any formula for sum of inverse trigonometric function now what we see here is we see sum of three sine inverse functions and we know that principal value range of any sine inverse function is from minus pi by 2 to plus pi by 2 so this sine inverse x will be less than equal to pi by 2 this sine inverse y will also be less than equal to pi by 2 and this sine inverse z

20:53 will also be less than or equal to pi by 2 but on the right hand side we have 3 pi by 2 which is pi by 2 plus pi by 2 plus pi by 2 so that means this right hand side is only possible when this sine inverse x is pi by 2 this sine inverse y is pi by 2 and sine inverse z is also pi by 2 so we'll take sine inverse of x is pi by 2 we'll take the sine on the right hand side we'll write x as sine pi by 2 and sine 90 is 1. so value of x is 1 in the same way value of y is 1 and the value of z is 1. so this is nothing but 1 plus 1 plus 1 plus 1 plus 1 plus 1 which is 6. so

21:39 answer to this question is simply 6. another question is cos inverse a1 plus cos inverse a2 plus cos inverse a3 plus cos inverse a n is 0 find the value of a 1 into a 2 into a 3 up to a n now again for this question if we look at cos inverse x principle value range of cos inverse x is from 0 to pi so cos inverse function is either 0 or a positive number now sum of all these cos inverses they cannot take a negative value so their sum can only be 0 if all of them

22:27 they're 0. so from here it will write cos inverse a i it is 0 then a i will be cos 0 and cos 0 is nothing but 1 so each of this a i is 1. so a1 is 1 a2 is 1 a3 is 1 a n is 1 so the value of a1 into a2 into a3 into a n will be simply 1 and that is the answer to this question now let us take another example which is if sine inverse a square plus cos inverse b square plus secant inverse c square plus cosec inverse d square is 5 pi square by 2 then the value of sine inverse a square minus cos inverse b square plus secant inverse c square minus cosec inverse d now when we say sine inverse a square we know that

23:13 maximum value of sine inverse is pi by 2 so sine inverse a square will always be less than equal to pi square by 4 in the same way cos inverse b square will always be less than or equal to pi square secant inverse c whole square will be less than equal to pi square and cosec inverse d square will always be less than or equal to pi square by 4. now we'll add them all up we'll get sine inverse a square plus cos inverse b square plus secant inverse c square plus cosec inverse d square and it'll be less

24:01 than or equal to 2 pi square plus pi square by 4 so it'll be 5 pi square by 2. now in the question we are given this equality sign so this equality sign is possible if and only if all these values they are equal so from here we'll get sine inverse a as plus or minus pi by 2 so value of a will be plus or minus sine pi by 2 which is plus or minus 1 cos inverse b it will be plus or minus pi now it cannot take the value minus pi lb plus pi

24:47 now cos pi is -1 so value of d will be minus 1 in the same way secant inverse c lb pi so c will be minus 1 and cosec inverse d will be again plus or minus pi by 2 value of d will be plus or minus 1. and the question we are not supposed to find the value of abcd i've just done that to show you how do we find the values of abcd we just need to find the value of this expression now sine inverse a square is pi square by 4 so this value is pi square by 4 minus pi squared plus pi square and then minus

25:33 pi square by 4 which is nothing but 0 and that is the answer to this question now there'll be questions where we'll need to use domain of trigonometric inverse functions also so let us take a function which is f x is sine inverse log to the base 2 x by 2 and we need to find domain of this function now for sine inverse we know that its domain is from minus 1 to plus 1 so we can write this log x by 2 to the base 2 it must lie between minus 1 and plus 1. now we'll take anti-law since base is greater than 1 it is an increasing function sine won't

26:18 change so we'll get this as 2 to the power -1 is less than equal to x by 2 is less than equal to 2 now we'll multiply it with 2 so we'll get 1 is less than equal to x is less than equal to 4 so domain of this function is from 1 to 4. now there'll be many questions in trigonometric inverse functions which can be solved just by looking at domain of the function say for example this question which says find the number of real solutions of 10 inverse under root x into x plus 1 plus sine inverse under root x square plus x plus 1 equals pi by 2. now what we'll do is we'll first find domain of this

27:04 question now for this 10 inverse x we know that domain of 10 inverses x belongs to r and then we have the square root for the square root condition should be x into x plus 1 it should be greater than or equal to 0. now for this sine inverse we know that its domain is from minus 1 to plus 1 and since it is a square root it cannot take a negative value so from here we'll get this condition that under root of x square plus x plus 1 it should be less than or equal to 1. now we'll square it we'll get x square plus x plus 1 is less than or equal to 1 and the condition will be x square plus x should be less than or equal to zero

27:49 so we're getting two simultaneous conditions one is this x into x plus one should be zero or greater than zero and here it is x into x plus one should be zero or less than zero now both of them they can only be simultaneously true if and only if x into x plus one is zero so the domain of this function is just two values zero and 1 so for this question there are only two values of x which are possible and they're 0 and minus 1 now we'll take x as 0 here will be 10 inverse 0 plus sine inverse one ten inverse zero is zero and sine inverse one is pi by two

28:35 we'll get this value pi by two so x equal to zero is one solution and it will take x is minus one now again it'll be 10 inverse 0 and plus sine inverse 1 which again is pi by 2 so that means there are only two real solutions of this equation and that's your option number c now the next question is if the domain of this function f x equals under root 3 cos inverse 4 x minus pi is a to b then the value of 4 a plus 64 b is now we know that for this under root function 3 cos inverse 4x minus pi it should be

29:22 greater than or equal to 0 so from here we'll write cos inverse 4x it should be greater than or equal to pi by 3 now here we have this cos inverse function and we know that principal value range of cross inverse function is from 0 to pi so additionally we have this condition that cos inverse 4x must be less than or equal to pi now this cos inverse function if we recall its graph it is a decreasing function so when we take cos the sign of inequality will change so we'll get cos pi by 3 it'll be less than or equal to 4x and it'll be greater than or equal to cos pi now cos pi by 3 is 1 by 2

30:10 and it'll be greater than or equal to 4x and will be greater than or equal to minus 1 so value of x it should lie between minus 1 by 4 and 1 by 8. so here the value of a is minus 1 by 4 and the value of b is 1 by 8. now we need to find the value of 4 a plus 64 b now value of a is minus 1 by 4 and value of b is 1 by 8 so it will be minus 1 plus 8 and that is 7. so answer to this question will be simply 7. now let us take another question sine inverse x square plus 2 x plus 2 plus 10 inverse x square minus 3x minus

30:56 k square is greater than pi by 2 we need to find the value of k now we can write this as sine inverse and here it will be x plus 1 whole square plus 1 plus 10 inverse x square minus 3x minus k square and it is greater than pi by 2. now we know the domain of sine inverse x is x should lie between minus 1 and plus 1 and domain of 10 inverse x is x belongs to r so we don't have a problem with 10 inverse so for the sine inverse we can write x plus one whole square plus one it should lie between minus one and plus one now this is always true because the right hand side is always positive now

31:42 we'll solve this condition now one and one will cancel we get x plus 1 whole square it is less than or equal to 0. now square of a number cannot be negative so the only possible condition in this case is value of x should be -1 so for this inequality there is only one value of x which is possible so domain of this question will be value of x should be minus 1. now we'll put x as minus 1 we'll get sine inverse 1 plus 10 inverse now lb 1 plus 3 minus k square and it is greater than pi by 2 now sine inverse 1

32:30 is pi by 2 so this pi by 2 and pi by 2 will cancel so we'll get this condition that 10 inverse 4 minus k square is greater than 0 now 10 inverse it is an increasing function so it will take 10 both sides we get 4 minus k square it should be greater than 0 that is k square is less than 4 or mod k is less than 2 which means value of k should lie between minus 2 and plus 2 which satisfies all the 4 options so answer is a b c and d now another question is the trigonometric equation sine inverse x equals 2 sine inverse a has a solution for which of the following conditions of a

33:16 now we know that this equation will have a solution if the value on the right hand side lies in its principal value range so it simply means this 2 sine inverse a it must lie between minus pi by 2n plus pi by 2. so we connect minus pi by 4 is less than equal to sine inverse a is less than equal to pi by 4. now we'll take sine then sine minus pi by 4 will be minus 1 by root 2 is less than equal to a is less than equal to 1 by root 2. so from here we get this condition that mod of a is less than or equal to 1 by root 2 and that's your option d now

34:03 here the question is we are given two sets e 1 and e 2 and we are given two functions f x and g x f x is log x upon x minus 1 and g x is sine inverse log x upon x minus 1. now first we will find domain of fng so we'll start with this function gx for this function gx we have sine inverse function and for sine inverse domain is x should lie between minus 1 and plus 1. so for this function gx we can write this log e x upon x minus one it should lie between minus one and plus one now base is greater than 1 it will take

34:48 antilog sine won't change so we get 1 upon e is less than x upon x minus 1 is less than e now we'll subtract 1 and add 1 we can write 1 upon e is less than 1 plus 1 upon x minus 1 is less than e or we can write 1 minus e upon e is less than equal to 1 upon x minus 1 and it is less than e minus 1. now this interval it contains 0. now we have this condition that 1 upon x lies between a and b where 0 belongs to a and b so here either the value of x is less than 1 upon a

35:35 or the value of x is greater than 1 upon b so from here we'll get this condition that x minus 1 it should be less than or equal to e upon 1 minus e that is value of x is less than equal to e upon 1 minus e plus 1 or x minus 1 is greater than equal to 1 upon e minus 1 that is x is greater than equal to 1 upon e minus 1 plus 1 so from here we get this condition that x is less than or equal to 1 upon 1 minus e or x is greater than or equal to e upon e minus 1 so domain of this function g which is e2 it is defined as

36:21 minus infinity to 1 upon 1 minus e inclusive union e to e minus 1 to infinite and that's your option 1 so this s it matches with 1 now we'll find domain of this function f now domain of this function f is e 1 and even is the condition that x upon x minus 1 is greater than 0 now this is 0 and 1 so it'll be plus minus n plus so we'll get the condition that either x is less than 0 or x is greater than 1 and it is given that for e2 x belongs to e1 that is e2 is a

37:09 subset of e1 so domain of g which is e2 it is contained in domain of f so domain of f will contain domain of g that means this r also it matches with 1. now we look at all other options which contains the element either less than 0 or greater than 1 so between 0 and 1 it is not possible it contains element between 0 and 1 it will also have elements between 0 and 1 again it's not possible and it is also not possible so this r it also matches with only s now we need to find a range of f and range of g now this function f it is defined as

37:58 log x upon x minus 1 let us take this as y we get x upon x minus 1 is e to the power y we'll get x minus 1 as e to the power minus y into x so x will be 1 upon 1 minus e to the power minus y now what is the restriction for this function restriction for this function is denominator it must not be zero so we get e to the power minus y it is unequal to one that is value of pi should not be zero so range of f it will be y belongs to minus infinity to 0 union 0 to infinite so we'll take all the values except 0 so range of f is

38:44 this option 4 so this p it matches with 4 now we'll come to this function gx and our gx is actually sine inverse of fx now range of fx is either fx is less than 0 or fx is greater than 0 so we'll take fx less than 0 or f x is greater than zero if we take sine inverse we'll get sine inverse f x is less than sine inverse zero which is zero and in the same way sine inverse f x is greater than sine inverse zero so this gx it is either less than 0 or

39:31 gx is greater than 0 and since gx is sine inverse function we know that its principal value range will lie between minus pi by 2 and plus pi by 2 so range of this gx will be either from minus pi by 2 to not inclusive 0 union 0 to inclusive pi by 2 and the only option which is contained in this set is this option 2. so range of g contains the set 2 so q it matches with 2. so answer to this question is p matches with 4 q matches with 2 r with 1 and s with 1 and that's your option a and that is the answer to this question

40:17 in this video we'll be proving properties of inverse trigonometric functions now we have already seen in lecture 1 that any inverse trigonometric function is nothing but an angle say for example if we write sine inverse 1 by 2 value of sine is 1 by 2 at pi by 6 so sine inverse 1 by 2 is nothing but an angle whose value is pi by 6 in the principal value range of sine inverse now we can write the same angle in terms of other trigonometric ratios also now we know that cos pi by 6 is root 3 by 2 so we can also express this angle pi by 6 as cos inverse

41:04 root 3 by 2 we also know that 10 30 is 1 by root 3 so we can also write this as 10 inverse 1 by root 3. so by looking at this we can infer the conclusion that it is possible to change sine inverse to cos inverse or tan inverse or any other trigonometric inverse function so the first thing before we will start proving the properties is how to change one inverse trigonometric function to another now suppose we have sine inverse x and we want to change it to cos inverse so what we'll do is we'll take the sine inverse excess some angle theta because any trigonometric inverse function is nothing but an angle now we can take sine on the right hand

41:49 side we can write x as sine theta now once we have sine theta we need to write cos theta now how do we write cos theta there are two ways we know that cos theta is under root of 1 minus sine square theta now sine theta is x so we can write 1 minus x square so we can write this theta as cos inverse of under root 1 minus x square so we can change sine inverse x as cos inverse under root of 1 minus x square now there will be some restriction on the domain of x when will be possible but as of now we are not going into the

42:35 detail we'll study the detail once we have studied self-adjusting property in inverse trigonometric functions so as of now we are dealing with the basic method to change one trigonometric inverse function into another now another way of conversion is we can take a right angle triangle this is sine theta so it'll be x and this is 1 and lb 1 minus x square what is cos theta and this cos theta will be 1 minus x square so we can simply write theta as cos inverse under root of 1 minus x square so sine inverse x is broadly equal to cos inverse under root 1 minus x square now we have to change it to 10 we can write 10 theta

43:22 and 10 theta in this case is x upon under root 1 minus x square so we can write this as 10 inverse x upon under root 1 minus x square now we can do the same for any trigonometric expression suppose we are given cos inverse 2 by 3 and we have to change it to sine inverse and 10 inverse so what we'll do is we'll take this as theta so we'll have cos theta as 2 by 3 now we'll take a right angle triangle this is theta this is 2 and this is 3 and here will be 9 minus 4 under root 5.

44:09 so we can write this as sine inverse now what is sine thetas antidotes root 5 by 3 so it will be sine inverse under root 5 by 3 and what is tan theta 10 theta in this case is root 5 by 2 so we'll write this as 10 inverse under root 5 by 2. so this is how we are going to change one trigonometric inverse function into another now the first property is sine inverse of minus x it is minus sine inverse of x now how do we prove it so what we'll do is we'll let sine inverse of minus x as some angle theta so we'll get

44:56 minus x equals sine theta now we can take this minus on the right hand side we can write x as minus sine theta now we know that sine minus theta is minus sine theta so we can actually write this as x equals sine minus theta now again we'll take inverse function so we can write minus theta equals sine inverse x or theta is minus sine inverse of x so sine was minus x equals theta and theta equals minus sine inverse x so from here we can say sine inverse of minus x is equal to minus sine inverse x so if we have

45:42 sine inverse minus 1 by root 2 we can write this as minus sine inverse 1 by root 2 and sine inverse 1 by root 2 is pi by 4 so it'll be minus pi by 4. now the second one is cos inverse of minus x now cos inverse minus x is pi minus cos inverse x now for this what we'll do is we'll again take this as theta so we'll write cos inverse of minus x as theta so we'll write minus x equals cos theta or x equals minus cos theta now here cos minus theta is not minus cos theta but cos pi minus theta is minus cos theta so we can write

46:29 x as cos pi minus theta now we take inverse function so we can write cos inverse x equals pi minus theta or theta is pi minus cos inverse of x now comparing these two we can write cos inverse minus x equals pi minus cos inverse of x so we have to write cos inverse of minus 1 by 2 we'll have to write this as pi minus cos inverse 1 by 2 so it'll be pi minus now what is cos inverse 1 by 2 cos inverse 1 by 2 is pi by 3 so we'll be 2 pi by 3. now if you'll recall principal value range of sine inverse

47:15 and cos inverse so principle value range of sine inverses minus pi by 2 to plus pi by 2 so if sine has a negative value the angle should be in fourth quadrant and we are getting this angle in the fourth quadrant and for cos inverse principal value ranges from 0 to pi so if cos is taking a negative value the answer must lie in second quadrant which is true in this case so this is how we prove properties for sine inverse minus x and cos inverse minus x i'll just write the result for other expressions you can prove them the same way so third one is 10 inverse minus x and 10 inverse minus x is

48:01 minus 10 inverse x now cosec inverse x it is same as sine inverse so cosec inverse minus x will be minus cosec inverse of x secant inverse of minus x lb same as that of course will be pi minus secant inverse of x and make a note for this one and which is cot inverse of minus x cot inverse of minus x is not same as 10 inverse of minus x so it is not minus cot inverse x it is actually pi minus cot inverse of x and the reason for that

48:48 is principal value range for 10 inverses minus pi by 2 to plus pi by 2 and principle value range for cot inverse axis from 0 to pi so 10 versus negative in fourth quadrant whereas cot inverse is negative in second quadrant so we have to get the resultant second quadrant and that is the reason it is not minus cot inverse x but pi minus cot inverse of x now we come to second set of properties and which is sine inverse of x is equal to cosec inverse of 1 by x so we'll just take the reciprocal of the

49:34 number we can change sine inverse into cosec inverse now how do we prove it now this again is simple so let sine inverse x s theta then x will be simply sine theta now we'll take reciprocal 1 upon sine theta is cosec theta and it'll be equal to 1 by x so we can write this as theta equals cosec inverse 1 by x so from here we can say sine inverse x is equal to cosec inverse 1 by x and in the same way we can write cos inverse x as secant inverse 1 by x and the third one is

50:21 tan inverse xs quad inverse 1 by x and here in the third one we need to remember that this is true only when x is greater than 0 because when x is a negative number there is a difference in quadrant for 10 inverse and cot inverse so tan inverse x is equal to cot inverse 1 by x only when x is a positive number so if we have secant inverse 2 we can write this as cos inverse 1 by 2 and cos inverse 1 by 2 is nothing but pi by 3 suppose if we have cot inverse minus root 3

51:07 now we cannot write this as 10 inverse minus 1 by root 3 because this formula it is only valid when x is greater than 0 so in this case what do we actually do so if it is quart inverse of minus root 3 now we know that cot inverse minus x is pi minus cot inverse x so we'll write this as pi minus cot inverse of root 3 now this number is positive now we can write cot inverse x as 10 1 by x so we'll write pi minus 10 inverse 1 by root 3 so it'll be pi minus and this is pi by 6 so it'll be 5 pi by 6. so that's your second set of properties now third set of properties is actually

51:54 trigonometric identities so for inverse trigonometric function we have three set of identities and the first one is sine inverse x plus cos inverse x and it is equal to pi by 2 where x belongs to minus 1 and plus 1 that is it holds in the domain of sine inverse and cos inverse now how do we prove this result so again what we'll do is we'll let sine inverse x sum theta we'll get xs sine theta now we know that cos 90 minus theta is sine theta so we can write x as cos

52:40 pi by 2 minus theta so we can write inverse xs pi by 2 minus theta now we'll take this theta on the left hand side and theta is sine inverse x so from here we can write sine inverse x plus cos inverse x and it is equal to pi by 2 so this result it is valid for any value of x which lies between minus 1 and plus 1 so it is an identity in inverse trigonometric function in the same way we can prove tan inverse x plus quad inverse x equals pi by 2 when x belongs to r

53:26 and third one is secant inverse of x plus cosec inverse of x equals pi by 2 when x belongs to minus infinity to minus 1 union 1 to infinite now you get some simple questions such as y equals sine inverse x square upon x square plus 1 plus secant inverse x square plus 1 upon x square find d y by dx now what we know is secant inverse of x is cos inverse 1 by

54:12 x so we can write this as y equals sine inverse x square upon x square plus 1 plus cos inverse x square upon x square plus 1 now here this value is same so it is sine inverse x plus cos inverse x and which is nothing but pi by 2 so this function is actually a constant function so we write d y by dx d y by dx in this case is nothing but zero so derivative of this function is nothing but zero now we'll come to two of the most important properties of trigonometric inverse function now we have already studied in function that if f and f inverse are inverse functions of each other then

54:57 f of f inverse is equal to f inverse of f and this is equal to x so self-adjusting property in inverse trigonometric function mean that sine of sine inverse x is equal to x when x lies in the domain of sine inverse when x lies between minus 1 and plus 1 in the same way cos of cos inverse x is equal to x when x lies between -1 and plus 1 10 of 10 inverse x is also x when x belongs to r

55:44 cot of cot inverse x is x when x belongs to r secant of secant inverse x is x when x belongs to minus infinity to minus 1 union 1 to infinite and cosec of cosec inverse x is equal to x when again x lies in minus 1 into minus 1 union 1 to infinite and this is very simple to prove so again in this we'll take sine inverse xs sum theta clearly x is sine theta now we'll take sine both sides we can write sine of

56:29 sine inverse x is nothing but sine theta and sine theta is nothing but x so we can write sine of sine inverse x it is equal to x so basically f and f inverse they'll cancel out each other and we'll get this x provided this x lies in the domain of the function so if we have sine of sine inverse 2 by 3 we'll cancel this sine and sine inverse we can write this result as 2 by 3 but if we have sine of sine inverse 3 by 2 now we cannot cancel it we cannot write this as 3 by 2 because this sine of 3 by 2 in itself is not defined so all these self-existing properties they're very simple and they

57:14 should be used considering the domain of any inverse trigonometric function now there'll be questions on other self-adjusting property and which is cos of course inverse x is x and sine of sine inverse x is x if x lies in the domain of respective inverse trigonometric functions now here the question is if x lies between 0 and 1 then under root of 1 plus x square x cos cot inverse x plus sine cot inverse x square minus 1 to the power 1 by 2 is equal to which of the following now we'll take this as y so this is y equals under root of 1 plus x square and then it is x

58:00 cos cot inverse x plus sine cot inverse of x square minus 1 whole to the power 1 by 2. now what we'll do is here we'll let cot inverse x is theta so it'll be got inverse x equals theta so x equals cot theta now we'll replace x with cot theta we can write y equals under root of 1 plus cot square theta and lb

58:46 cot theta into now this is cos theta plus sine theta whole square minus 1 whole to the power 1 by 2 so we can write this as y equals now 1 plus cot square theta is cosec square theta so it will be cosec theta and here it will be cos square theta plus sine square theta upon sine theta whole square minus 1 to the power 1 by 2 now cos square theta plus sine square theta it is 1 so we'll write this as cosec theta and here will be cosec

59:31 square theta minus 1 to the power 1 by 2 now because x square theta minus 1 is cot square theta so we can write this as y equals cosec theta into cot theta so this y it simplifies to cos theta upon sine square theta now we have taken cot theta as x so we'll draw a right angle triangle now this is theta this is x this is one and it'll be at the root of one plus x square now cos theta is x upon under root of

1:00:17 1 plus x square and sine square theta will be 1 upon 1 plus x square so it simplifies to x into under root of 1 plus x square and that's your option c now let us take a matching type problem so we'll start with this p which is we need to simplify one upon y square cos of tan inverse y plus y of sine of tan inverse y upon cot of

1:01:04 sine inverse y plus 10 of sine inverse y whole square plus y to the power 4 whole to the power 1 by 2 and let us take this expression as p now first we'll simplify this expression in the numerator which is cos of tan inverse y plus y sine tan inverse of i now here we let tan inverse of y as theta so we can write 10 theta equals y

1:01:50 so we'll draw a right angle triangle this is why this is 1 and this is under root of 1 plus y square so this expression it simply becomes now cos theta plus y sine theta now cos theta is 1 upon under root of 1 plus y square and then y and sine theta is y upon under root of 1 plus y square so we'll simplify we can simply write this as under root of 1 plus y square now we'll simplify this expression in the denominator which is

1:02:37 cot of sine inverse y plus 10 of sine inverse y now here we let sine inverse y is theta so we can write sine theta equals y so this is theta this is why this is 1 and it'll be 1 minus y square now we need to find the value of cot theta plus tan theta now cot theta here is under root of

1:03:23 1 minus y square upon y and plus 10 theta is y upon under root of 1 minus y square so it'll be 1 upon y into under root of 1 minus y square now we'll put these two values in the expression and we'll find the value of p so we can write this p as 1 upon y square into now cos tan inverse y plus y sine 10 y which is under root of 1 plus y square upon then this value 1 upon y

1:04:11 1 minus y square whole square plus y to the power 4 whole to the power 1 by 2. now here y square and y square will cancel so we'll get this p as under root of 1 minus y to the power 4 whole square plus y to the power 4 to the power 1 by 2. now square and square root will cancel y to the power 4 will also cancel so value of p will be simply 1 so this p it matches with this option 4 so p it matches with 4. now the second part is if cos x plus cos y plus cos z equals

1:04:57 zero and sine x plus sine y plus sine z is also zero then the possible value of cos x minus y by two is now here what we'll do is we'll write x plus cos y is minus cos z and sine x plus sine y is minus sine z now we'll square both the equations and then we'll add them so we'll write cos x plus cos y whole square plus sine x plus sine y whole square and it'll be cos square z plus sine square z now here cos square x plus sine square x will be 1

1:05:43 in the same way cos square y plus sine square by lb1 plus 2 and then we'll get cos x cos y plus sine x sine y and this value will be one so this one and one will cancel now this is cos a cos b minus sine a sine b which is nothing but cos a minus b so we can write cos x minus y is minus 1 by 2 now using half angle formula we can write cos x minus y is 2 cos square x minus y by 2 minus 1 now value of cos x minus y is nothing but minus 1 by 2.

1:06:30 so from here we can write 2 cos square x minus y by 2 lb 1 by 2 or cos x minus y by 2 will be plus or minus 1 by 2 which matches with this option 3 so q it matches with 3. now we'll come to the third part which is r and it says if cos pi by 4 minus x into cos 2x plus sine x into sine 2x into secant x is equal to

1:07:17 cos x into sine 2x into secant x plus cos pi by 4 plus x into cos 2x we need to find one of the possible values of secant x now what we'll do is we'll take these two terms together and we'll take these other two terms together so we can write cos 2x into cos pi by 4 minus x minus cos pi by 4 plus x and it'll be equal to sine 2x into secant x

1:08:04 and it'll be cos x minus sine x we can simplify this as cos 2x and this is cos a cos b plus sine a sine b minus cos a cos b minus sine a sine b so we'll get two sine pi by 4 into sine x and here we get 2 sine x into cos x into secant x and lb cos x minus sine x we can also write this cos 2x is

1:08:50 cos square x minus sine square x now here we can cancel cos x minus sine x so one of the solution that we'll get is cos x minus sine x equals 0 we can also cancel 2 sine x so another solution that we'll get is sine x equals 0 here cos x into secant x will be 1 so now we are left with cos x plus sine x into 1 by root 2 and it'll be equal to 1. so from here we'll get this condition that cos x plus sine x

1:09:38 and it'll be root 2. now from this first condition we'll get x as pi by 4 from the second condition we'll get x as 0 and from this third condition we'll get x is pi by 4 so here the value of secant x will be secant 0 which is 1 or secant pi by 4 which is root 2. so this r it matches with either 4 or 2 but there's no 4 given in this column so we'll take r matches with 2 so p is 4 q is 3 r is 2

1:10:24 and that's your option b but we'll still verify this s is cot of sine inverse under root of 1 minus x square it is sine of 10 inverse root 6x now since it involves sine inverse under root of 1 minus x square clearly x should lie between minus 1 and plus 1. now first we'll simplify this expression on the left hand side which is

1:11:09 cot of sine inverse of under root of 1 minus x square so we'll take this as theta we can write sine theta is under root of 1 minus x square so this is your theta and this is under root of 1 minus x square this is 1 then lbx now we have to write cot theta cot theta will be x upon under root of 1 minus x square now we'll simplify this right hand side now this rhs is sine of tan inverse of

1:11:55 x root 6 now we'll take this as some angle theta so here tan theta is under root 6x so this is theta this is root 6x and this is 1 and this is 1 plus 6x square now we'll write sine theta sine theta will be under root 6 x upon under root of 1 plus 6 x square now we rewrite this equation we can write this equation as x into under root of 1 minus x square it is equal to under root of 6 x upon under root of

1:12:42 1 plus 6 x square now here x and x will cancel so one of the value of x will be x equals 0 and then we'll square it we'll get 6 minus 6 x square and it'll be 1 plus 6 x square so x square will be simply 5 by 12 or the value of x is 1 by 2 under root 5 by 3 so one of the possible value of x is 1 by 2 under root 5 by 3 which is this option 1 so this s it matches with one so p4 q3 r2 and s1 and that's your option b [Music]