Question

If $\alpha ={\sin }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)$ and $\beta ={\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)$, then $\tan \alpha \cdot \tan \beta$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

We have, $\alpha ={\sin }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)$
$={\sin }^{-1}\left(\cos (\frac{\pi }{2}-{\cos }^{-1}x)\right)={\sin }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)$
and $\beta ={\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)$
$\therefore \alpha +\beta ={\sin }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)+{\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)$
$=\frac{\pi }{2}$
Now, consider
$\tan \alpha \tan \beta =\tan \alpha \cdot \tan (\frac{\pi }{2}-\alpha )$
$=\tan \alpha \cdot \cot \alpha =1$