Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

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

Solution:

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

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

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

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

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

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

$=\mathrm{RHS}$

Hence proved.

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