Question

If $f : R \rightarrow\left[\frac{-\pi}{4}, \frac{\pi}{2}\right)$ is a function defined as $f ( x )=\tan ^{-1}\left( x ^2-2 x +\cos ^{-1} \alpha-\frac{3 \pi}{2}+\sin ^{-1} \beta\right)$ is a surjective function then

• A. $\sin \left(\sin ^{-1}(\alpha+\beta)\right)=\cos \left(\cos ^{-1}(\alpha+\beta)\right)$
• B. $\sin ^{-1}(\sin (\alpha-\beta))=\cos ^{-1}(\cos (\alpha-\beta))$
• C. $\tan ^{-1}\left(\tan \left(\alpha^2+\beta^2\right)\right)=\cot ^{-1}\left(\cot \left(\alpha^2+\beta^2\right)\right)$
• D. $\tan \left(\tan ^{-1}\left(\alpha^2-\beta^2\right)\right)=\cot \left(\cot ^{-1}\left(\alpha^2-\beta^2\right)\right)$

Answer 1

Answer: (A), (D)

$\because f ( x )$ is surjective function
$\therefore \frac{-\pi}{4} \leq f ( x )<\frac{\pi}{2}$
$\because x^2-2 x+\cos ^{-1} \alpha-\frac{3 \pi}{2}+\sin ^{-1} \beta=(x-1)^2+\cos ^{-1} \alpha+\sin ^{-1} \beta-\frac{3 \pi}{2}-1$
for surjective
$\cos ^{-1} \alpha+\sin ^{-1} \beta-\frac{3 \pi}{2}-1=-1 \Rightarrow \cos ^{-1} \alpha+\sin ^{-1} \beta=\frac{3 \pi}{2}$
$\therefore \cos ^{-1} \alpha=\pi$ and $\sin ^{-1} \beta=\frac{\pi}{2} \Rightarrow \alpha=-1$ and $\beta=1$
(A) $ \therefore \sin \left(\sin ^{-1}(\alpha+\beta)\right)=0=\cos \left(\cos ^{-1}(\alpha+\beta)\right)$
(B) $ \sin ^{-1}(\sin (\alpha-\beta))=\sin ^{-1}(\sin (-2))=-(\pi-2)=2-\pi$
But $\cos ^{-1}(\cos (\alpha-\beta))=\cos ^{-1}(\cos (-2))=2$
(C) $\tan ^{-1}\left(\tan \left(\alpha^2+\beta^2\right)\right)=\tan ^{-1}(\tan 2)=2-\pi$ But $\cot ^{-1}\left(\cot \left(\alpha^2+\beta^2\right)\right)=\cot ^{-1}(\cot 2)=2$
(D) $ \tan \left(\tan ^{-1}(0)\right)=0$ and $\cot \left(\cot ^{-1}(0)\right)=0 . $