Question

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

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

Answer 1

Answer: (A)

$2ta{n}^{-1}\left[\sqrt{\frac{a-b}{a+b}}tan\frac{\theta }{2}\right]$
$=co{s}^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right)ta{n}^2\frac{\theta }{2}}{1+\left(\frac{a-b}{a+b}\right)ta{n}^2\frac{\theta }{2}}\right]$
$\left[\because 2ta{n}^{-1}x=co{s}^{-1}\frac{1-x^2}{1+x^2}\right]$
$=co{s}^{-1}\left[\frac{\left(a+b\right)-\left(a-b\right)ta{n}^2\frac{\theta }{2}}{\left(a+b\right)+\left(a-b\right)ta{n}^2\frac{\theta }{2}}\right]$
$=co{s}^{-1}\left[\frac{a\left(1-ta{n}^2\frac{\theta }{2}\right)+b\left(1+ta{n}^2\frac{\theta }{2}\right)}{a\left(1+ta{n}^2\frac{\theta }{2}\right)+b\left(1-ta{n}^2\frac{\theta }{2}\right)}\right]$
$=co{s}^{-1}\left[\frac{\frac{a\left(1-ta{n}^2\frac{\theta }{2}\right)}{1+ta{n}^2\frac{\theta }{2}}+b}{a+b\left(\frac{1-ta{n}^2\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}}\right)}\right]$
$=co{s}^{-1}\left[\frac{acos\theta +b}{a+bcos\theta }\right]$