#### Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 9 maths Textbook Solution.

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

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

Hint: Using $2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

Solution: Let us assume

\begin{aligned} \text { L.H.S } &=2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right] \\ &=\cos ^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right] \end{aligned}                                        $\left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$

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

\begin{aligned} &=\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\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right] \end{aligned}

Dividing numerator and denominator by $\left(1+\tan ^{2} \frac{\theta}{2}\right)$ we get

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

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

$= R.H.S$

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