Inverse Trigonometric Functions Miscellaneous NCERT Solutions

1. Find the value of

\cos^{-1}\left(\cos\frac{13\pi}{6}\right)

Solution

We have to find the value of

\cos^{-1}\left(\cos\frac{13\pi}{6}\right)

Now,

\frac{13\pi}{6}=2\pi+\frac{\pi}{6}

Using the periodic property of cosine function,

\cos\left(2\pi+\theta\right)=\cos\theta

Therefore,

\cos\frac{13\pi}{6}=\cos\frac{\pi}{6}

Hence,

\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\cos\frac{\pi}{6}\right)

Now, the principal value branch of \cos^{-1}x is

[0,\pi]

Since

\frac{\pi}{6}\in[0,\pi]

therefore,

\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}

Hence,

\boxed{\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\frac{\pi}{6}}

2. Find the value of

\tan^{-1}\left(\tan\frac{7\pi}{6}\right)

Solution

We have to find the value of

\tan^{-1}\left(\tan\frac{7\pi}{6}\right)

Now,

\frac{7\pi}{6}=\pi+\frac{\pi}{6}

Using the periodic property of tangent function,

\tan(\pi+\theta)=\tan\theta

Therefore,

\tan\frac{7\pi}{6}=\tan\frac{\pi}{6}

Hence,

\tan^{-1}\left(\tan\frac{7\pi}{6}\right)=\tan^{-1}\left(\tan\frac{\pi}{6}\right)

Now, the principal value branch of \tan^{-1}x is

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Since

\frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

therefore,

\tan^{-1}\left(\tan\frac{\pi}{6}\right)=\frac{\pi}{6}

Hence,

\boxed{\tan^{-1}\left(\tan\frac{7\pi}{6}\right)=\frac{\pi}{6}}

3. Prove that

2\sin^{-1}\frac35=\tan^{-1}\frac{24}{7}

Solution

In such questions involving different inverse trigonometric functions, an easy way is to convert one function into another by using the basics of trigonometry using a right angled triangle. Check the solution:

We have to prove that

2\sin^{-1}\frac35=\tan^{-1}\frac{24}{7}

Let

\theta=\sin^{-1}\frac35

Then,

\sin\theta=\frac35

Since the principal value branch of \sin^{-1}x is

\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

and \sin\theta=\frac35>0, therefore

\theta lies in the first quadrant.

Now, consider a right triangle with:

• Perpendicular =3
• Hypotenuse =5

Then by Pythagoras theorem,

\text{Base}=\sqrt{5^2-3^2}=\sqrt{25-9}=4

Thus,

\tan\theta=\frac34

Now, using the double angle formula,

\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}

Substituting \tan\theta=\frac34,

We get, \tan2\theta=\frac{2\times\frac34}{1-\left(\frac34\right)^2}

=\frac{\frac32}{1-\frac{9}{16}}

=\frac{\frac32}{\frac{7}{16}}

or

=\frac32\times\frac{16}{7}

=\frac{24}{7}

Therefore,

\tan2\theta=\frac{24}{7}

Taking \tan^{-1} on both sides, we get

2\theta=\tan^{-1}\frac{24}{7}

But,

\theta=\sin^{-1}\frac35

Therefore,

2\sin^{-1}\frac35=\tan^{-1}\frac{24}{7}

Hence proved.

4. Prove that

\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35=\tan^{-1}\frac{77}{36}

Solution

We have to prove that

\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35=\tan^{-1}\frac{77}{36}

In questions involving the sum of inverse trigonometric functions, a good approach is to assign the inverse functions as angles and then use suitable trigonometric identities. Since the RHS contains a single inverse tangent term, it is natural to use the tan’s addition formula.

Firstly, Let

\alpha=\sin^{-1}\frac{8}{17}

and

\beta=\sin^{-1}\frac35

Then,

\sin\alpha=\frac{8}{17}

and

\sin\beta=\frac35

Since the principal value branch of \sin^{-1}x is

\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

and both sine values are positive, therefore

\alpha and \beta both lie in the first quadrant.

Now, for \alpha, consider a right triangle with:

• Perpendicular =8
• Hypotenuse =17

Then,

\text{Base}=\sqrt{17^2-8^2}=\sqrt{289-64}=15

Thus,

\tan\alpha=\frac{8}{15}

Similarly, for \beta, consider a right triangle with:

• Perpendicular =3
• Hypotenuse =5

Then,

\text{Base}=\sqrt{5^2-3^2}=\sqrt{25-9}=4

Therefore,

\tan\beta=\frac34

Now,

\alpha+\beta=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35

Using the formula

\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}

We get, \tan(\alpha+\beta)=\frac{\frac{8}{15}+\frac34}{1-\frac{8}{15}\times\frac34}

=\frac{\frac{32+45}{60}}{1-\frac{24}{60}}

=\frac{\frac{77}{60}}{\frac{36}{60}}

i.e.

=\frac{77}{36}

Therefore,

\tan(\alpha+\beta)=\frac{77}{36}

Taking \tan^{-1} on both sides,

\alpha+\beta=\tan^{-1}\frac{77}{36}

Substituting the values of \alpha and \beta, we get

\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35=\tan^{-1}\frac{77}{36}

Hence proved.

5. Prove that

\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}

Solution

We have to prove that

\cos^{-1}\frac45+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}

In questions involving the sum of inverse trigonometric functions, a good approach is to assign the inverse functions as angles and then use suitable trigonometric identities. Since the RHS contains a single inverse cosine term, it is natural to use the cosine addition formula.

Let

\alpha=\cos^{-1}\frac45

and

\beta=\cos^{-1}\frac{12}{13}

Then,

\cos\alpha=\frac45

and

\cos\beta=\frac{12}{13}

Since the principal value branch of \cos^{-1}x is

[0,\pi]

and both cosine values are positive, therefore

\alpha and \beta both lie in the first quadrant.

Now, for \alpha, consider a right triangle with:

• Base =4
• Hypotenuse =5

Then by Pythagoras theorem,

\text{Perpendicular}=\sqrt{5^2-4^2}=\sqrt{25-16}=3

Therefore,

\sin\alpha=\frac35

Similarly, for \beta, consider a right triangle with:

• Base =12
• Hypotenuse =13

Then,

\text{Perpendicular}=\sqrt{13^2-12^2}=\sqrt{169-144}=5

Thus,

\sin\beta=\frac5{13}

Now, using the cosine addition formula

\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta

we get, \cos(\alpha+\beta)=\frac45\times\frac{12}{13}-\frac35\times\frac5{13}

=\frac{48}{65}-\frac{15}{65}

=\frac{33}{65}

Therefore,

\cos(\alpha+\beta)=\frac{33}{65}

Now, both \alpha and \beta lie in the first quadrant, so their sum also lies in the principal value range of \cos^{-1}x. Hence, taking inverse cosine on both sides gives

\alpha+\beta=\cos^{-1}\frac{33}{65}

Substituting the values of \alpha and \beta, we get

\cos^{-1}\frac45+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}

Hence proved.

6. Prove that

\cos^{-1}\frac{12}{13}+\sin^{-1}\frac35=\sin^{-1}\frac{56}{65}

Solution

We have to prove that

\cos^{-1}\frac{12}{13}+\sin^{-1}\frac35=\sin^{-1}\frac{56}{65}

Whenever we see a sum involving inverse trigonometric functions, it is usually helpful to convert them into angles and then use standard trigonometric identities. Since the RHS contains a single inverse sine term, we should try to find the sine of the sum.

Let

\alpha=\cos^{-1}\frac{12}{13}

and

\beta=\sin^{-1}\frac35

Then,

\cos\alpha=\frac{12}{13}

and

\sin\beta=\frac35

Since the principal value branch of \cos^{-1}x is [0,\pi] and \cos\alpha>0, therefore \alpha lies in the first quadrant.

Also, the principal value branch of \sin^{-1}x is \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \sin\beta>0, therefore \beta also lies in the first quadrant.

Now, for \alpha, consider a right triangle with:

• Base =12
• Hypotenuse =13

Then by Pythagoras theorem,

\text{Perpendicular}=\sqrt{13^2-12^2}=\sqrt{169-144}=5

Therefore,

\sin\alpha=\frac5{13}

Similarly, for \beta, consider a right triangle with:

• Perpendicular =3
• Hypotenuse =5

Then,

\text{Base}=\sqrt{5^2-3^2}=\sqrt{25-9}=4

Thus,

\cos\beta=\frac45

Now, using the sine addition formula

\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta

We get, \sin(\alpha+\beta)=\frac5{13}\times\frac45+\frac{12}{13}\times\frac35

=\frac{20}{65}+\frac{36}{65}

=\frac{56}{65}

Therefore,

\sin(\alpha+\beta)=\frac{56}{65}

Now, both \alpha and \beta lie in the first quadrant, so \alpha+\beta lies in the principal value range of \sin^{-1}x. Hence, taking inverse sine on both sides, we get

\alpha+\beta=\sin^{-1}\frac{56}{65}

Substituting the values of \alpha and \beta, we get

\cos^{-1}\frac{12}{13}+\sin^{-1}\frac35=\sin^{-1}\frac{56}{65}

Hence proved.

7. Prove that

\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac35

Solution

We have to prove that

\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac35

In this question, the RHS is a sum of two inverse trigonometric functions, while the LHS is a single inverse tangent term. Hence, a natural approach is to assume the inverse functions as angles and then use the tangent addition formula.

Let

\alpha=\sin^{-1}\frac{5}{13}

and

\beta=\cos^{-1}\frac35

Then,

\sin\alpha=\frac{5}{13}

and

\cos\beta=\frac35

Since the principal value range of \sin^{-1}x is \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \sin\alpha>0, therefore \alpha lies in the first quadrant.

Also, the principal value range of \cos^{-1}x is [0,\pi] and \cos\beta>0, therefore \beta also lies in the first quadrant.

Now, for \alpha, consider a right triangle with:

• Perpendicular =5
• Hypotenuse =13

Then by Pythagoras theorem,

\text{Base}=\sqrt{13^2-5^2}=\sqrt{169-25}=12

Therefore,

\tan\alpha=\frac5{12}

Similarly, for \beta, consider a right triangle with:

• Base =3
• Hypotenuse =5

Then,

\text{Perpendicular}=\sqrt{5^2-3^2}=\sqrt{25-9}=4

Thus,

\tan\beta=\frac43

Now, using the tangent addition formula

\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}

We get, \tan(\alpha+\beta)=\frac{\frac5{12}+\frac43}{1-\frac5{12}\times\frac43}

=\frac{\frac{15+48}{36}}{1-\frac{20}{36}}

=\frac{\frac{63}{36}}{\frac{16}{36}}

i.e.

=\frac{63}{16}

Therefore,

\tan(\alpha+\beta)=\frac{63}{16}

Now, both \alpha and \beta lie in the first quadrant, so \alpha+\beta lies in the principal value range of \tan^{-1}x. Hence, taking inverse tangent on both sides, we get

\alpha+\beta=\tan^{-1}\frac{63}{16}

Substituting the values of \alpha and \beta, we get

\sin^{-1}\frac{5}{13}+\cos^{-1}\frac35=\tan^{-1}\frac{63}{16}

Hence proved.

8. Prove that

\tan^{-1}\sqrt{x}=\frac12\cos^{-1}\frac{1-x}{1+x},\quad x\in[0,1]

Solution

We have to prove that

\tan^{-1}\sqrt{x}=\frac12\cos^{-1}\frac{1-x}{1+x},\quad x\in[0,1]

In this question, the RHS contains a factor of \frac12, which suggests that we should use a double angle identity. Since the inverse cosine term is present, the identity involving \cos2\theta will be most useful.

Let

\theta=\tan^{-1}\sqrt{x}

Then,

\tan\theta=\sqrt{x}

Since x\in[0,1], therefore

\sqrt{x}\ge0

Hence,

\theta\in\left[0,\frac{\pi}{4}\right]

Now, using the identity

\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}

Substituting \tan\theta=\sqrt{x}, we get

\cos2\theta=\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}

=\frac{1-x}{1+x}

Therefore,

2\theta=\cos^{-1}\frac{1-x}{1+x}

This is valid because

\theta\in\left[0,\frac{\pi}{4}\right]

therefore,

2\theta\in\left[0,\frac{\pi}{2}\right]

which lies in the principal value branch of \cos^{-1}x, namely [0,\pi].

Dividing throughout by 2, we get

\theta=\frac12\cos^{-1}\frac{1-x}{1+x}

But,

\theta=\tan^{-1}\sqrt{x}

Therefore,

\tan^{-1}\sqrt{x}=\frac12\cos^{-1}\frac{1-x}{1+x}

Hence proved.

9. Prove that

\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2},\quad x\in\left(0,\frac{\pi}{4}\right)

Solution

In this question, the expression inside \cot^{-1} contains square roots involving 1\pm\sin x. Such expressions are usually simplified using half-angle identities.

Recall that

1+\sin x=\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2

and

1-\sin x=\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2

Since

x\in\left(0,\frac{\pi}{4}\right)

therefore,

\frac{x}{2}\in\left(0,\frac{\pi}{8}\right)

Hence both \sin\frac{x}{2} and \cos\frac{x}{2} are positive.

Therefore,

\sqrt{1+\sin x}=\sin\frac{x}{2}+\cos\frac{x}{2}

and

\sqrt{1-\sin x}=\cos\frac{x}{2}-\sin\frac{x}{2}

Substituting these values in the given expression, we get

\cot^{-1}\left(\frac{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)+\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)-\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}\right)

=\cot^{-1}\left(\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)

=\cot^{-1}\left(\cot\frac{x}{2}\right)

Now, the principal value branch of \cot^{-1}x is

(0,\pi)

Since

\frac{x}{2}\in\left(0,\frac{\pi}{8}\right)\subset(0,\pi)

therefore,

\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}

Hence,

\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}

Hence proved.

10. Prove that

\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac12\cos^{-1}x,\quad -\frac{1}{\sqrt2}\le x\le1

Solution

In this question, the RHS contains \frac12\cos^{-1}x, which suggests that half-angle identities may be useful. Also, the complicated expression involving square roots can often be simplified by expressing them in terms of trigonometric functions.

Let

\theta=\cos^{-1}x

Then,

\cos\theta=x

Since the principal value branch of \cos^{-1}x is

[0,\pi]

and

-\frac{1}{\sqrt2}\le x\le1

therefore,

0\le\theta\le\frac{3\pi}{4}

Hence,

0\le\frac{\theta}{2}\le\frac{3\pi}{8}

Now, using the half-angle identities

1+\cos\theta=2\cos^2\frac{\theta}{2}

and

1-\cos\theta=2\sin^2\frac{\theta}{2}

So, we get, \sqrt{1+x}=\sqrt{1+\cos\theta}

=\sqrt{2\cos^2\frac{\theta}{2}}

=\sqrt2\cos\frac{\theta}{2}

Similarly, \sqrt{1-x}=\sqrt{1-\cos\theta}

=\sqrt{2\sin^2\frac{\theta}{2}}

=\sqrt2\sin\frac{\theta}{2}

Substituting these values in the given expression, we get

\tan^{-1}\left(\frac{\sqrt2\cos\frac{\theta}{2}-\sqrt2\sin\frac{\theta}{2}}{\sqrt2\cos\frac{\theta}{2}+\sqrt2\sin\frac{\theta}{2}}\right)

=\tan^{-1}\left(\frac{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}\right)

Dividing numerator and denominator by \cos\frac{\theta}{2}, we get

=\tan^{-1}\left(\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}\right)

Now, using the identity

\tan\left(\frac{\pi}{4}-A\right)=\frac{1-\tan A}{1+\tan A}

we get

=\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)

Now,

0\le\frac{\theta}{2}\le\frac{3\pi}{8}

or -\frac{3\pi}{8}\le\frac{\theta}{2}\le0

Now, adding \frac{\pi}{4} throughout, we get

-\frac{\pi}{8}\le\frac{\pi}{4}-\frac{\theta}{2}\le\frac{\pi}{4}

which lies in the principal value branch of \tan^{-1}x, namely

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Hence,

\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)=\frac{\pi}{4}-\frac{\theta}{2}

Therefore,

\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{\theta}{2}

But,

\theta=\cos^{-1}x

Hence,

\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac12\cos^{-1}x

Hence proved.

11. Solve the following equation

2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)

Solution

We have to solve the equation

2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)

In equations involving inverse trigonometric functions, a good strategy is to reduce the equation into a standard trigonometric equation by applying suitable identities.

Let

\theta=\tan^{-1}(\cos x)

Then,

\tan\theta=\cos x

The given equation becomes

2\theta=\tan^{-1}(2\cosec x)

Taking tangent on both sides, we get

\tan2\theta=2\cosec x

Now, using the identity

\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}

and substituting \tan\theta=\cos x, we get

\frac{2\cos x}{1-\cos^2x}=2\cosec x

Using

1-\cos^2x=\sin^2x

we get

\frac{2\cos x}{\sin^2x}=2\cosec x

=\frac{2}{\sin x}

Therefore,

\frac{2\cos x}{\sin^2x}=\frac{2}{\sin x}

Multiplying both sides by \sin^2x, we get

2\cos x=2\sin x

\cos x=\sin x

Dividing both sides by \cos x, we get

\tan x=1

Now, we know that

\tan\frac{\pi}{4}=1

Also, the tangent function has period \pi. This means its value repeats after every interval of \pi.

Therefore, if

\tan x=1

then possible values of x are

\frac{\pi}{4},\ \pi+\frac{\pi}{4},\ 2\pi+\frac{\pi}{4},\ -\pi+\frac{\pi}{4},\ \text{etc.}

Hence, in general, we can write

x=n\pi+\frac{\pi}{4},\quad n\in\mathbb{Z}

Therefore, the required solution is

\boxed{x=n\pi+\frac{\pi}{4},\quad n\in\mathbb{Z}}

12. Solve the following equation

\tan^{-1}\frac{1-x}{1+x}=\frac12\tan^{-1}x,\quad (x>0)

Solution

Since the RHS contains a factor of \frac12, this suggests that we should use the tangent half-angle identity.

Let

\theta=\tan^{-1}x

Then,

\tan\theta=x

Since x>0, therefore

\theta\in\left(0,\frac{\pi}{2}\right)

The given equation becomes

\tan^{-1}\frac{1-\tan\theta}{1+\tan\theta}=\frac{\theta}{2}

Now, using the identity

\tan\left(\frac{\pi}{4}-A\right)=\frac{1-\tan A}{1+\tan A}

we get

\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right)=\frac{\theta}{2}

Now, since

\theta\in\left(0,\frac{\pi}{2}\right)

therefore,

\frac{\pi}{4}-\theta\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)

which lies in the principal value branch of \tan^{-1}x, namely

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Hence,

\frac{\pi}{4}-\theta=\frac{\theta}{2}

Multiplying throughout by 2, we get

\frac{\pi}{2}-2\theta=\theta

\frac{\pi}{2}=3\theta

So, \theta=\frac{\pi}{6}

But,

\theta=\tan^{-1}x

Therefore,

x=\tan\frac{\pi}{6}

=\frac{1}{\sqrt3}

Hence, the required solution is

\boxed{x=\frac{1}{\sqrt3}}

13. \sin(\tan^{-1}x),\ |x|<1 is equal to

• (A) \frac{x}{\sqrt{1-x^2}}
• (B) \frac{1}{\sqrt{1-x^2}}
• (C) \frac{1}{\sqrt{1+x^2}}
• (D) \frac{x}{\sqrt{1+x^2}}

Solution

In questions involving expressions like \sin(\tan^{-1}x), a very useful approach is to assume the inverse trigonometric expression as an angle and then use a right triangle to find the required trigonometric ratio.

Let

\theta=\tan^{-1}x

Then,

\tan\theta=x=\frac{x}{1}

Now, consider a right triangle with:

• Perpendicular =x
• Base =1

Then by Pythagoras theorem,

\text{Hypotenuse}=\sqrt{x^2+1^2}=\sqrt{1+x^2}

Therefore,

\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{x}{\sqrt{1+x^2}}

But,

\theta=\tan^{-1}x

Hence,

\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}

✅️ Therefore, the correct answer is

(D) \frac{x}{\sqrt{1+x^2}}

14. \sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}, then x is equal to

• (A) 0,\ \frac12
• (B) 1,\ \frac12
• (C) 0
• (D) \frac12

Solution

In equations involving inverse trigonometric functions, a good approach is to substitute the inverse expression by an angle and then simplify using standard trigonometric identities.

Let

\sin^{-1}x=\theta

Then,

x=\sin\theta

The given equation becomes

\sin^{-1}(1-\sin\theta)-2\theta=\frac{\pi}{2}

So, we get

\sin^{-1}(1-\sin\theta)=\frac{\pi}{2}+2\theta

Taking sine on both sides,

1-\sin\theta=\sin\left(\frac{\pi}{2}+2\theta\right)

Using

\sin\left(\frac{\pi}{2}+A\right)=\cos A

we get

1-\sin\theta=\cos2\theta

Now, using the identity

\cos2\theta=1-2\sin^2\theta

we get

1-\sin\theta=1-2\sin^2\theta

2\sin^2\theta-\sin\theta=0

\sin\theta(2\sin\theta-1)=0

Hence,

\sin\theta=0

or

\sin\theta=\frac12

Since x=\sin\theta, the possible values are

x=0

or

x=\frac12

Now, we must check which values actually satisfy the original equation.

For x=0,

\sin^{-1}(1-0)-2\sin^{-1}(0)

=\sin^{-1}(1)-0

=\frac{\pi}{2}

Hence, x=0 satisfies the equation.

Now, for x=\frac12,

\sin^{-1}\left(1-\frac12\right)-2\sin^{-1}\left(\frac12\right)

=\sin^{-1}\left(\frac12\right)-2\left(\frac{\pi}{6}\right)

=\frac{\pi}{6}-\frac{\pi}{3}

=-\frac{\pi}{6}

which does not satisfy the equation.

✅️ Therefore, the only value satisfying the equation is

(C) 0