tan-1 \(\big[\frac{\cos\,X}{1+\sin\,X}\big]\), x\(\big[\frac{-\pi}{2},\frac{\pi}{2}\big]\)
tan-1\(\big[\frac{\cos\,X}{1+\sin\,X}\big]\) \(\because\)\(\Bigg\{\begin{matrix}
\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} \\
and\,1+\sin x = \big(\cos\frac{x}{2}+\sin \frac{x}{2}\big)^2
\end{matrix}\)
= tan-1\(\Bigg[\frac{\cos^2\frac{X}{2}-\sin^2\frac{X}{2}}{\big(\cos\frac{X}{2}+\sin\frac{X}{2}\big)^2}\Bigg]\)
= tan-1\(\Bigg[\frac{\big(\cos\frac{X}{2}+\sin \frac{X}{2}\big)\big(\cos\frac{X}{2}-\sin\frac{X}{2}\big)}{\big(\cos \frac{X}{2}+\sin\frac{X}{2}\big)^2}\Bigg]\)
= tan-1\(\Bigg[{\cos\frac{X}{2}-\sin\frac{X}{2} \over\cos\frac{X}{2}+\sin\frac{X}{2}}\Bigg]\) Divide by cos\(\frac{X}{2}\), we get
= tan-1 \(\Bigg[{1-\tan\frac{X}{2} \over 1+\tan\frac{X}{2}}\Bigg]\)
= tan-1\(\Big[\tan\Big(\frac{\pi}{4}-\frac{X}{2}\Big)\Big]\) = \(\frac{\pi}{4}-\frac{X}{2}\)
