Prove the following identities (1-16)

Question:

Prove the following identities (1-16)

$\frac{1-\sin x \cos x}{\cos x(\sec x-\operatorname{cosec} x)} \cdot \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{3} x+\cos ^{3} x}=\sin x$

Solution:

LHS $=\frac{1-\sin x \cos x}{\cos x(\sec x-\operatorname{cosec} x)} \times \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{3} x+\cos ^{3} x}$

$=\frac{1-\sin x \cos x}{\cos x\left(\frac{1}{\cos x}-\frac{1}{\sin x}\right)} \times \frac{(\sin x)^{2}-(\cos x)^{2}}{(\sin x)^{3}+(\cos x)^{3}}$

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

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

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

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

$=\sin x$

$=$ RHS

Hence proved.

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