Question

What is the real values of x which satisfy function $:\frac{({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3}{({\tan }^{-1}x+{\cot }^{-1}x{)}^3}=7$

• A. $x=-1$
• B. $x=+1$
• C. $x=-2$
• D. $x=+2$

Answer 1

The correct option is A $x=-1$

$\frac{({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3}{({\tan }^{-1}x+{\cot }^{-1}x{)}^3}=7$

We know, ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$

$\Rightarrow ({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=\frac{7{\pi }^3}{8}$ (i)

Now Let $y=({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=({\sin }^{-1}x+{\cos }^{-1}x)\left[\right({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2-{\sin }^{-1}x.{\cos }^{-1}x]$

$\Rightarrow y=\frac{\pi }{2}\left[\right({\sin }^{1-}x+{\cos }^{-1}x{)}^2-3{\sin }^{-1}x.{\cos }^{-1}x]$

$\Rightarrow y=\frac{\pi }{2}[\frac{{\pi }^2}{4}-3{\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)]$

$\Rightarrow y=\frac{\pi }{2}[\frac{{\pi }^2}{16}+3{(\frac{\pi }{4}-{\sin }^{-1}x)}^2]=\frac{7{\pi }^3}{8}$ using (i)

After simplification $\Rightarrow {\sin }^{-1}x=-\frac{\pi }{2}\Rightarrow x=-1$