Question

$2{\tan }^{-1}\left[\sqrt{\left(\frac{a-b}{a+b}\right)t}\mathrm{an}\left(\frac{\theta }{2}\right)\right]=$

• A. ${\cos }^{-1}\frac{[a\cos \theta +b]}{[a+b\cos \theta ]}$
• B. ${\cos }^{-1}\frac{[a+b\cos \theta ]}{[a\cos \theta +b]}$
• C. ${\cos }^{-1}\left[\frac{a\cos \theta }{[a+b\cos \theta ]}\right]$
• D. ${\cos }^{-1}\left[\frac{b\cos \theta }{[a\cos \theta +b]}\right]$

Answer 1

The correct option is A

${\cos }^{-1}\frac{[a\cos \theta +b]}{[a+b\cos \theta ]}$

Explanation for the correct option:

Find the value of given expression:

Given, $2{\tan }^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}\right)$

$={\cos }^{-1}\left(\frac{1-{\displaystyle \frac{a-b}{a+b}}{\tan }^2\frac{\theta }{2}}{1-{\displaystyle \frac{a-b}{a+b}}{\tan }^2\frac{\theta }{2}}\right)$ $\left[\mathbf{\because }\mathbf{2}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{;}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{<}\mathit{x}\mathbf{<}\mathbf{1}\right]$

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

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

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

$\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{[}\mathbf{a}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{]}\mathbf{}}{\mathbf{[}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }\mathbf{]}}$ $\left[\mathbf{\because }\frac{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}{\displaystyle \frac{\mathbf{\theta }}{\mathbf{2}}}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\frac{\mathbf{\theta }}{\mathbf{2}}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right]$

Hence, Option ‘A’ is Correct.