Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

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

Solution:

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

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

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

$=\left(\frac{\cos ^{2} x}{1-\cos ^{4} x}+\frac{\sin ^{2} x}{1-\sin ^{4} x}\right) \sin ^{2} x \cos ^{2} x$

$=\left(\frac{\cos ^{2} x\left(1-\sin ^{4} x\right)+\sin ^{2} x\left(1-\cos ^{4} x\right)}{\left(1-\cos ^{4} x\right)\left(1-\sin ^{4} x\right)}\right) \sin ^{2} x \cos ^{2} x$

$=\left(\frac{1-\cos ^{2} x \sin ^{4} x-\cos ^{4} x \sin ^{2} x}{\left(1+\sin ^{2} x\right)\left(1-s \operatorname{in}^{2} x\right)\left(1+\cos ^{2} x\right)\left(1-\cos ^{2} x\right)}\right) \sin ^{2} x \cos ^{2} x$

$=\left(\frac{1-\cos ^{2} x \sin ^{4} x-\cos ^{4} x \sin ^{2} x}{\left(1+\sin ^{2} x\right) \cdot \cos ^{2} x \cdot\left(1+\cos ^{2} x\right) \cdot \sin ^{2} x}\right) \sin ^{2} x \cos ^{2} x$

$=\left(\frac{1-\cos ^{2} x \sin ^{4} x-\cos ^{4} x \sin ^{2} x}{\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)}\right)$

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

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

= RHS

Hence peoved.