Question

The value of $2{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}\right]$ is equal to

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

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{a\cos \theta +b}{a+b\cos \theta }\right)$

$2{\tan }^{-1}[\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}]$

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

$[\because 2{\tan }^{-1}x={\cos }^{-1}\frac{1-x^2}{1+x^2}]$

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

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

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

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