Question
Prove that $\tan^{–1 }\Bigg(\frac{\text{\cos x}}{\text{1 + sinx}}\Bigg)=\frac{\pi}{4}-\frac{x}{2},\text{x}\in\Bigg(-\frac{\pi}{\text{2}},\frac{{\pi}}{2}\Bigg).$

Answer

$\tan^{-1}\Bigg(\frac{\cos\text{x}}{\text{1+sinx}}\Bigg)=\tan^{-1}\Bigg(\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{\text{1}+\cos\Big(\frac{\pi}{2}-\text{x}\Big)}\Bigg)$
$=\tan^{-1}\Bigg(\frac{2\sin\Big(\frac{\pi}{2}-\frac{\text{x}}{2}\Big)\cos\Big(\frac{\pi}{4}-\frac{\text{x}}{2}\Big)}{2\cos^{2}\Big(\frac{\pi}{2}-\frac{\text{x}}{2}\Big)}\Bigg)=\tan^{-1}\Bigg(\tan\frac{\pi}{4}-\frac{\text{x}}{2}\Bigg)$
$=\frac{\pi}{4}-\frac{\text{x}}{2}$.

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