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  4. The value 2 tan-1 [√(a-b/a+b) tan(θ/2)] is equal to

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

Inverse Trigonometric Functions

Solution:

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