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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions2 Marks
Question
Express $\tan ^{-1}( \frac{\cos x}{1-\sin x}),-\frac{3 \pi}{2}<x<\frac{\pi}{2}$ in the simplest form.
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Answer

According to question, we have 
$\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ = $\tan ^{-1}\left[\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}\right]$
= $\tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$
= $\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2}$

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