Question

The value of $\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]$ is equal to

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

Answer 1

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

Let ${\cos }^{-1}\frac{a}{b}=2\theta \Rightarrow \cos 2\theta =\frac{a}{b}$

$\Rightarrow \tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]=\tan [\frac{\pi }{4}+\theta ]$

$=\frac{1+\tan \theta }{1-\tan \theta }=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }=\frac{1+\sin 2\theta }{\cos 2\theta }$

$=1+\frac{\frac{\sqrt{b^2-a^2}}{b}}{\frac{a}{b}}=\frac{b+\sqrt{b^2-a^2}}{a}$