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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
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 $\tan ^{-1}\left\{\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \frac{x}{2}\right\}=\theta$
$\therefore \tan \theta =\sqrt{\frac{\alpha-\beta}{\alpha+\beta}} \tan \cdot \frac{x}{2}$
now $\text{L.H.S.}  =2 \theta=\cos ^{-1}(\cos 2 \theta)$
$ =\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]$
Putting the values :$=\cos ^{-1}\left(\frac{1-\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}{1+\frac{\alpha-\beta}{\alpha+\beta} \tan ^2 \frac{x}{2}}\right)$
$=\cos ^{-1}\left(\frac{(\alpha+\beta) \cos ^2 \frac{x}{2}-(\alpha-\beta) \sin ^2 \frac{x}{2}}{(\alpha+\beta) \cos ^2 \frac{x}{2}+(\alpha-\beta) \sin ^2 \frac{x}{2}}\right)$
$=\cos ^{-1}\left(\frac{\beta\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\alpha\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}{\alpha\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+\beta\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}\right)$
$=\cos ^{-1}\left(\frac{\beta+\alpha \cos x}{\alpha+\beta \cos x}\right)=\text { R.H.S. }$

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