Question

Which one of the following is correct?

• A. $\sin \left({\cos }^{-1}x\right)=\cos \left({\sin }^{-1}x\right)$
• B. $\sec \left({\tan }^{-1}x\right)=\tan \left({\sec }^{-1}x\right)$
• C. $\cos \left({\tan }^{-1}x\right)=\tan \left({\cos }^{-1}x\right)$
• D. $\tan \left({\sin }^{-1}x\right)=\sin \left({\tan }^{-1}x\right)$

Answer 1

The correct option is A

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

Explanation for the correct option

For option (a)

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

From the properties of trigonometry with a right-angle triangle, we get

If $△ABC$is a right-angle triangle, then

$$\begin{gathered} \sin \alpha =\frac{AB}{AC} \\ \cos \alpha =\frac{BC}{AC} \\ \tan \alpha =\frac{AB}{BC} \end{gathered}$$

L.H.S.

$\sin \left({\cos }^{-1}x\right)$

Let ${\cos }^{-1}x=\theta$

$\Rightarrow x=\cos \theta$

Therefore, $\sin \theta =\frac{\sqrt{1-x^2}}{1}$………………(1)

R.H.S.

$\cos \left({\sin }^{-1}x\right)$

Let ${\sin }^{-1}x=\varphi$

$\Rightarrow x=\sin \varphi$

Therefore, $\cos \varphi =\frac{\sqrt{1-x^2}}{1}$………………(2)

From (1) and (2), we get L.H.S. = R.H.S.

Explanation for incorrect options

For option (b)

$\sec \left({\tan }^{-1}x\right)$

Let ${\tan }^{-1}x=\theta$

$\Rightarrow x=\tan \theta$

Therefore, $\sec \theta =\frac{\sqrt{1+x^2}}{1}$………………(3)

$\tan \left({\sec }^{-1}x\right)$

Let $se{c}^{-1}x=\varphi$

$\Rightarrow x=sec\varphi$

Therefore, $\tan \varphi =\frac{\sqrt{1-x^2}}{1}$………………(4)

From (3) and (4), we get L.H.S. $\ne$ R.H.S.

For option (c)

$\cos \left({\tan }^{-1}x\right)$

Let ${\tan }^{-1}x=\theta$

$\Rightarrow x=\tan \theta$

Therefore, $\cos \theta =\frac{1}{\sqrt{1+x^2}}$………………(5)

$\tan \left({\cos }^{-1}x\right)$

Let ${\cos }^{-1}x=\varphi$

$\Rightarrow x=\cos \varphi$

Therefore, $\tan \varphi =\frac{\sqrt{1-x^2}}{x}$………………(6)

From (5) and (6), we get L.H.S. $\ne$ R.H.S.

For option (d)

$\tan \left({\sin }^{-1}x\right)$

Let ${\sin }^{-1}x=\theta$

$\Rightarrow x=\sin \theta$

Therefore, $\tan \theta =\frac{x}{\sqrt{1-x^2}}$………………(7)

$\sin \left({\tan }^{-1}x\right)$

Let ${\tan }^{-1}x=\varphi$

$\Rightarrow x=\tan \varphi$

Therefore, $\sin \varphi =\frac{x}{\sqrt{1+x^2}}$………………(8)

From (7) and (8), we get L.H.S. $\ne$ R.H.S.

Hence, the correct option is option (A).