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Question 15.5.5: Prove that cos x/(1-sin x)=tan (π/4+x/2)....

Prove that cos⁡x(1−sin⁡x)=tan⁡(π4+x2) \frac{\cos x}{(1-\sin x)}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right) .

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We have

 LHS =cos⁡x(1−sin⁡x) =(cos⁡2x2−sin⁡2x2)(cos⁡2x2+sin⁡2x2−2sin⁡x2cos⁡x2) [∵ cos⁡x=(cos⁡2x2−sin⁡2x2),cos⁡2x2+sin⁡2x2=1 and sin⁡x=2sin⁡x2cos⁡x2] \begin{array}{l} \text { LHS }=\frac{\cos x}{(1-\sin x)} \\  \\=\frac{\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)}{\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \\  \\\left[ \begin{array}{l} \because  \cos x=\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right), \cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}=1\\ \text{ and } \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \end{array} \right] \end{array} \\  \\

=(cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)(cos⁡x2−sin⁡x2)2 =(cos⁡x2+sin⁡x2)(cos⁡x2−sin⁡x2)=(1+tan⁡x2)(1−tan⁡x2)=(tan⁡π4+tan⁡x2)(1−tan⁡π4⋅tan⁡x2) [ dividing num. and denom. by cosx2] =tan⁡(π4+x2)= RHS. \begin{array}{l} =\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} }\\  \\=\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}=\frac{\left(1+\tan \frac{x}{2}\right)}{\left(1-\tan \frac{x}{2}\right)}=\frac{\left(\tan \frac{\pi}{4}+\tan \frac{x}{2}\right)}{\left(1-\tan \frac{\pi}{4} \cdot \tan \frac{x}{2}\right)} \\  \\ [ \text{ dividing num. and denom. by cos} \frac{x}{2} ]\\  \\=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\text { RHS. }\end{array}

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