Question

If $\sin \left[a{\tan }^{-1}\left\{\sqrt{\frac{b-x}{1+x}}\right\}\right]=\sqrt{1-x^c}$.
Find the value of a,b and c.

• A. $a=1,b=1,c=1$
• B. $a=1,b=2,c=2$
• C. $a=2,b=1,c=1$
• D. $a=2,b=1,c=2$

Answer 1

The correct option is D $a=2,b=1,c=2$

Given, $\sin \left[a{\tan }^{-1}\left\{\sqrt{\frac{b-x}{1+x}}\right\}\right]=\sqrt{1-x^c}$

Let $b=1$. and $x=\cos 2\theta$.

Then we get

$\sin \left[a{\tan }^{-1}\left\{\sqrt{\frac{b-x}{1+x}}\right\}\right]$

$=\sin \left[a{\tan }^{-1}\right(\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\left)\right]$
$=\sin \left[a{\tan }^{-1}\right(\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\left)\right]$
$=\sin \left[a{\tan }^{-1}\right(tan\theta \left)\right]$
$=\sin \left(a\theta \right)$

And $\sqrt{1-x^c}=\sqrt{1-{\cos }^{c}2\theta }$

$⟹\sin \left(a\theta \right)=\sqrt{1-{\cos }^{c}2\theta }$

Squaring both sides, we get

${\sin }^{2}a\theta =1-{\cos }^{c}2\theta$

${\cos }^{c}2\theta =1-{\sin }^{2}a\theta$

${\cos }^{c}2\theta ={\cos }^{2}a\theta$ ...... $[sin²x+cos²x=1]$

Hence $c=2$ and $a=2$.

Therefore $(a,b,c)=(2,1,2)$.