Question

Prove the following identities

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

 

Answer 1

\[\left( \frac{1}{\sec^2 x - \cos^2 x} + \frac{1}{{cosec}^2 x - \sin^2 x} \right) \sin^2 x \cos^2 x = \frac{1 - \sin^2 x \cos^2 x}{2 + \sin^2 x \cos^2 x}\]
\[\begin{array}{rcl}\text{ LHS } = & \left( \frac{1}{\sec^2 x - \cos^2 x} + \frac{1}{{cosec}^2 x - \sin^2 x} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{1}{\frac{1}{\cos^2 x} - \cos^2 x} + \frac{1}{\frac{1}{\sin^2 x} - \sin^2 x} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{\cos^2 x}{1 - \cos^4 x} + \frac{\sin^2 x}{1 - \sin^4 x} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{\cos^2 x\left( 1 - \sin^4 x \right) + \sin^2 x\left( 1 - \cos^4 x \right)}{\left( 1 - \cos^4 x \right)\left( 1 - \sin^4 x \right)} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{1 - \cos^2 x \sin^4 x - \cos^4 x \sin^2 x}{\left( 1 + \sin^2 x \right)\left( 1 - s {in}^2 x \right)\left( 1 + \cos^2 x \right)\left( 1 - \cos^2 x \right)} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{1 - \cos^2 x \sin^4 x - \cos^4 x \sin^2 x}{\left( 1 + \sin^2 x \right) . \cos^2 x . \left( 1 + \cos^2 x \right) . \sin^2 x} \right) \sin^2 x \cos^2 x \\ \\ = & \left( \frac{1 - \cos^2 x \sin^4 x - \cos^4 x \sin^2 x}{\left( 1 + \sin^2 x \right)\left( 1 + \cos^2 x \right)} \right) \\ \\ = & \frac{1 - \cos^2 x \sin^2 x\left( \sin^2 x + \cos^2 x \right)}{2 + \sin^2 x . \cos^2 x} \\ \\ = & \frac{1 - \cos^2 x \sin^2 x}{2 + \sin^2 x . \cos^2 x}\end{array}\]