Question

If $f:R\to \left[\frac{-\pi }{4},\frac{\pi }{2}\right)$ is a function defined as $f(x)={\tan }^{-1}\left(x^2-2x+{\cos }^{-1}\alpha -\frac{3\pi }{2}+{\sin }^{-1}\beta \right)$ 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
$\therefore \frac{-\pi }{4}\le f(x)<\frac{\pi }{2}$
$\because x^2-2x+{\cos }^{-1}\alpha -\frac{3\pi }{2}+{\sin }^{-1}\beta =(x-1{)}^2+{\cos }^{-1}\alpha +{\sin }^{-1}\beta -\frac{3\pi }{2}-1$
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 $\cot \left({\cot }^{-1}(0)\right)=0.$