Inverse Trigonometric Functions

Let us take = x

Therefore,

We know that principle value range of sin^-1 is

And,

Therefore principle value of is

Let us take

Then,

We know that principle value range of cos^-1 is [0, π]

And

Therefore, principle value of is

Let cosec^-1 2 = x

Therefore,

And

We know that principle value range of cosec^-1 is

Therefore, principle value of cosec^-1(2) is .

Let us take

Then we get,

And

We know that principle value range of tan^-1 is

Therefore, principle value of tan^-1 is.

Let us take

Then we will get,

And

We know that principle value range of cos^-1 is [0, π]

Therefore principle value of cos-1is

Let us take tan^-1(-1) = x then we get,

And

We know that principle value range of tan^-1 is

Therefore, principle value of tan^-1(-1) is

Let

Then,

We know that range of the principle value branch of sec^-1 is [0, π]

And

Therefore principle value of is

Let us consider

Then we get,

We know that the range of the principal value branch of cot^-1 is [0, π].

And,

Therefore, the principle value of is .

Let

Therefore,

We know that range of the principle value branch of cos^-1 is [0, π]

And

Therefore principle value of is

Let us take the values of

Then,

We know that range of the principle value branch of cosec^-1 is

And

Therefore principle value of is

Let us consider tan^-1(1) = x then we get

We know that range of the principle value branch of tan^-1 is

Therefore,

Let

We know that range of the principle value branch of cos^-1 is [0, π]

Therefore,

Let

We know that range of the principle value branch of sin^-1 is

Therefore

Now,

Let

Then, we get,

We know that range of the principle value branch of cos^-1 is [0, π]

Therefore

Let

We know that range of the principle value branch of sin^-1 is

Therefore

Now,

sin^–1 x = y

We know that range of the principle value branch of sin^-1 is

Therefore,

Hence, the option (B) is correct.

Let us take

Then we get,

We know that range of the principle value branch of tan^-1 is

Therefore,

Let Then we get,

We know that range of the principle value branch of sec^-1 is [0, π]

Therefore,

Now,

Hence, the option (B) is correct.

Let

We have,

R.H.S =

= sin^-1(sin (3θ))

= 3 θ

= 3sin^-1x

= L.H.S

Hence Proved

Let x = cos θ

Then, Cos^-1 = θ

Now, R.H.S. = cos^-1(4x³ - 3x)

= cos^-1(4cos³θ - 3cos θ)

= cos^-1(cos3θ)

= 3θ

= 3cos^-1x

= L.H.S.

Hence Proved

L.H.S.

= R.H.S.

Hence Proved.

L.H.S. =

= R.H.S.

Hence Proved.

Now, Put x = tanθ ⇒ θ =tan^-1x

Therefore,

Therefore,

Let us take,

[We have done this substitution on the bases of identity sec²θ – 1 = tan²θ]

Therefore,

Hence,

Dividing by cos x,

As we know tan(Π/4) = 1

Hence,

We will solve this problem on the bases of the identity 1 - sin²θ = cos²θ

So, for a² - x², we can substitute x = a sinθ or x = a cosθ

Now, Let us put x = a sinθ

Therefore,

Hence,

Put

Now,

= 3θ

We will solve the inner bracket first.

So, we will first find the principal value of

We know that,

Therefore,

= tan^-11

= π/4

Hence,

The value of

= 0

Hence, the value of cot(tan^-1a + cot^-1 a) = 0

We will solve this problem by expressing sin2θ and cos2θ in terms of tanθ

Now let us put, x = tanθ. Then we will have,

θ = tan^-1x

Now again, Let’s put, y = tan∅. Then we will have,

∅ = tan^-1y

Now,

Hence, the value of

On comparing the co-efficient on both sides we get,

(For type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range

So here,

Now, can be written as,

Hence, .

(For type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range.)

So here,

Now, can be written as,

Hence,

Let and

, so all ratio of y are positive and

and

Also,

as

So,

Hence,

(For cos^-1(cos x) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range.)

So here,




Now, can be written as,

where [since, cos(π + x) = -cosx]

as cos^-1(-x) = π – cos^-1

Hence, .

as

as

We all know that the principal value branch of sin^-1 is

Therefore,

Hence, the value of

(Forcos^-1(cosx) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range.)

So here,

Now, can be written as,

where

Hence,

(For tan^-1(tanx) type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range.)

So here,

Now, can be written as,

where, [since, tan(π+x) = tanx]

Hence,

Taking LHS

Let

Then,

Therefore,

……. (1)

and

As, LHS = RHS

Hence Proved !

………………………. (1)

Let

Then,

………………………. (2)

Now,

L.H.S.

Putting the value from equation (1) and (2))

=R.H.S.

Hence Proved.

Let

Then,

……………………….(1)

Let

Then,

……………(2)

Let

Then,

…………… (3)

Now,

L.H.S.

Putting the value from the equation (1) And (2))

……….by equation (3)

=R.H.S.

Hence Proved.

We can also solve this problem by using the identity Sin(A + B) = sinA cosB + cosA sinB

Letand

As R.H.S. is sin^-1 we will use sin (A+B)

= R.H.S.

Hence Proved.

Let Then,

LetThen,

Now,

R.H.S. =

Putting the value from equation (1) and (2))

=L.H.S.

Hence Proved.

L.H.S.

= R.H.S.

Hence Proved.

Let Then,

So now putting the value, we get,

R.H.S.

= L.H.S.

Consider,

On Rationalizing, we get,

Now,

L.H.S.

= R.H.S.

Hence Proved

Letso that

Now,

L.H.S.

= R.H.S.

Hence Proved.

Now, L.H.S.

Now, Let

L.H.S.

= R.H.S.

Hence Proved

Hence,

As we know,

We know,



So,

Hence,

Let then

Now we will put Now, we will put x= siny in the given equation, and we get,

⟹ 1 – siny = cos2y as

⟹ 1 - cos2y = sin y

⟹ 2sin²y = sin y

⟹ siny(2siny - 1) = 0

⟹ siny = 0 or 1/2

∴ x = 0 or 1/2

Now, if we put then we will see that,

L.H.S. =

R.H.S

Hence, is not the solution of the given equation.

Thus, x = 0