# Prove the following identities (1-16)

Question:

Prove the following identities (1-16)

$\frac{\cos x}{1-\sin x}=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$

Solution:

RHS $=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$

$=\frac{(1+\cos \mathrm{x})+(\sin \mathrm{x})}{(1+\cos \mathrm{x})-(\sin \mathrm{x})}$

$=\frac{[(1+\cos \mathrm{x})+(\sin \mathrm{x})][(1+\cos \mathrm{x})+(\sin \mathrm{x})]}{[(1+\cos \mathrm{x})-(\sin \mathrm{x})][(1+\cos \mathrm{x})+(\sin \mathrm{x})]}$

$=\frac{[(1+\cos x)+(\sin x)]^{2}}{(1+\cos x)^{2}-(\sin x)^{2}}$

$=\frac{(1+\cos x)^{2}+(\sin x)^{2}+2(1+\cos x)(\sin x)}{1^{2}+\cos ^{2} x+2 \times 1 \times \cos x-\sin ^{2} x}$

$=\frac{1+\cos ^{2} \mathrm{x}+2 \cos \mathrm{x}+\sin ^{2} \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}+2 \sin \mathrm{x}}{1+\cos ^{2} \mathrm{x}+2 \cos \mathrm{x}-\sin ^{2} \mathrm{x}}$

$=\frac{1+\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)+2 \cos \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}+2 \sin \mathrm{x}}{\left(1-\sin ^{2} \mathrm{x}\right)+\cos ^{2} \mathrm{x}+2 \cos \mathrm{x}}$

$=\frac{1+1+2 \cos x+2 \sin x \cos x+2 \sin x}{\cos ^{2} x+\cos ^{2} x+2 \cos x}$

$=\frac{2+2 \cos x+2 \sin x \cos x+2 \sin x}{2 \cos ^{2} x+2 \cos x}$

$=\frac{1+\cos x+\sin x \cos x+\sin x}{\cos ^{2} x+\cos x}$

$=\frac{1(1+\cos x)+\sin x(\cos x+1)}{\cos x(\cos x+1)}$

$=\frac{(\cos x+1)(1+\sin x)}{\cos x(\cos x+1)}$

$=\frac{(1+\sin x)}{\cos x}$

$=\frac{(1+\sin x) \times \cos x}{\cos x \times \cos x}$

$=\frac{(1+\sin x) \times \cos x}{\cos ^{2} x}$

$=\frac{(1+\sin x) \times \cos x}{1-\sin ^{2} x}$

$=\frac{(1+\sin x) \times \cos x}{(1+\sin x)(1-\sin x)}$

$=\frac{\cos x}{1-\sin x}$

$=\mathrm{LHS}$

Hence proved.

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