Prove that 2 ^ -1 - + x 2 = ^ -1 ( + x + x ) - Vidyadip

Question

Prove that $2 \tan ^{-1}\left\{\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \frac{x}{2}\right\}=\cos ^{-1}\left(\frac{\beta+\alpha \cos x}{\alpha+\beta \cos x}\right)$

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Answer

Suppose $\tan ^{-1}\left\{\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \frac{x}{2}\right\}=\theta$$
\begin{aligned}
\therefore \quad \tan \theta & =\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \cdot \frac{x}{2} \\
\text { now L.H.S. } & =2 \theta=\cos ^{-1}(\cos 2 \theta) \\
& =\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\left[\because \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]
\end{aligned}
$
Putting the values :$
=\cos ^{-1}\left(\frac{1-\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}{1+\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}\right)
$
$\begin{array}{l}=\cos ^{-1}\left(\frac{(\alpha+\beta) \cos ^2 \frac{x}{2}-(\alpha-\beta) \sin ^2 \frac{x}{2}}{(\alpha+\beta) \cos ^2 \frac{x}{2}+(\alpha-\beta) \sin ^2 \frac{x}{2}}\right) \\ =\cos ^{-1}\left(\frac{\beta\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\alpha\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}{\alpha\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\beta\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}\right) \\ =\cos ^{-1}\left(\frac{\beta+\alpha \cos x}{\alpha+\beta \cos x}\right)=\text { R.H.S. }\end{array}$

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