Question

The value of ${[\tan \{\frac{\pi }{4}+\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\}+\tan \{\frac{\pi }{4}-\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\}]}^{-1},$ where $0<a<b,$ is

• A. $\frac{b}{2a}$
• B. $\frac{a}{2b}$
• C. $\frac{\sqrt{b^2-a^2}}{2b}$
• D. $\frac{\sqrt{b^2-a^2}}{2a}$

Answer 1

The correct option is C $\frac{\sqrt{b^2-a^2}}{2b}$

Let $2\theta ={\sin }^{-1}\left(\frac{a}{b}\right)$
$\Rightarrow \sin 2\theta =\frac{a}{b}$
$\Rightarrow \cos 2\theta =\frac{\sqrt{b^2-a^2}}{b}$
Let $y=\tan \{\frac{\pi }{4}+\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\}+\tan \{\frac{\pi }{4}-\frac{1}{2}{\sin }^{-1}\left(\frac{a}{b}\right)\}$
$\Rightarrow y=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }\Rightarrow y=\frac{2(1+{\tan }^2\theta )}{(1-{\tan }^2\theta )}$
$\therefore \frac{1}{y}=\frac{1}{2}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)=\frac{1}{2}\cos 2\theta =\frac{\sqrt{b^2-a^2}}{2b}$