Question

Prove that:
(i) $2{\tan }^{-1}\sqrt{\frac{b}{a}}={\cos }^{-1}\left(\frac{a-b}{a+b}\right)$
(ii) Find the principal value of ${\cos }^{-1}(-\frac{1}{\sqrt{2}})$

Answer 1

(i) Let:

$\theta ={\tan }^{-1}\sqrt{\frac{b}{a}}\Rightarrow \sqrt{\frac{b}{a}}=\tan \theta$

Now, we use the trigonometric identity:

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

Substituting for $\theta$ and $\tan \theta$ and rearranging we get:

$\cos \left(2{\tan }^{-1}\sqrt{\frac{b}{a}}\right)=\frac{a-b}{a+b}$

$\Rightarrow 2{\tan }^{-1}\sqrt{\frac{b}{a}}={\cos }^{-1}\left(\frac{a-b}{a+b}\right)$

Hence Proved.

(ii) Let $y$ be the principal value of ${\cos }^{-1}(-\frac{1}{\sqrt{2}})$, then $0\le y\le {180}^o$

$\cos \left({135}^o\right)=-\frac{1}{\sqrt{2}}\Rightarrow y={135}^o$