Question

If ${\tan }^{-1}\left(\frac{\sqrt{a+x}-\sqrt{b-x}}{\sqrt{b+x}+\sqrt{a-x}}\right)=\frac{\pi }{c}-\frac{1}{d}{\cos }^{-1}x$, $-\frac{1}{\sqrt{2}}\le x\le 1$. Find the value of a,b,c and d

• A. $a=1,b=2,c=4,d=2$
• B. $a=2,b=2,c=2,d=1$
• C. $a=1,b=1,c=1,d=1$
• D. $a=1,b=1,c=4,d=2$

Answer 1

The correct option is D $a=1,b=1,c=4,d=2$

Let $a=b=k$ and $x=kcos2\theta$

Hence

$ta{n}^{-1}\left(\frac{\sqrt{a+x}-\sqrt{b-x}}{\sqrt{b+x}+\sqrt{a-x}}\right)$

$=ta{n}^{-1}\left(\frac{\sqrt{k+kcos2\theta }-\sqrt{k-kcos2\theta }}{\sqrt{k+kcos2\theta }+\sqrt{k-kcos2\theta }}\right)$

$=ta{n}^{-1}\left(\frac{\sqrt{kco{s}^2\theta }-\sqrt{ksi{n}^2\theta }}{\sqrt{kco{s}^2\theta }+\sqrt{ksi{n}^2\theta }}\right)$

$=ta{n}^{-1}\left(\frac{cos\theta -sin\theta }{cos\theta +sin\theta }\right)$

$=ta{n}^{-1}\left(\frac{1-\frac{sin\theta }{cos\theta }}{1+\frac{sin\theta }{cos\theta }}\right)$

$=ta{n}^{-1}\left(\frac{1-tan\theta }{1+tan\theta }\right)$

$=ta{n}^{-1}\left(tan\right(\frac{\pi }{4}-\theta \left)\right)$

$=\frac{\pi }{4}-\theta$

Now

$x=kcos2\theta$

Hence

$\theta =\frac{1}{2}co{s}^{-1}\left(\frac{x}{k}\right)$

Therefore

$\frac{\pi }{4}-\theta$$=$$\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\left(\frac{x}{k}\right)$

=$\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\left(\frac{x}{k}\right)$

$=\frac{\pi }{c}-\frac{1}{d}co{s}^{-1}\left(x\right)$

Hence by comparison we get

$c=4,d=2,k=1$

Since $a=b=k$.

Hence $a=b=1$.

Therefore $(a,b,c,d)=(1,1,4,2)$.