Question

Prove the following identities (1-16)
$\left(\frac{1}{{\sec }^{2}x-{\cos }^{2}x}+\frac{1}{{\mathrm{cosec}}^{2}x-{\sin }^{2}x}\right){\sin }^{2}x{\cos }^{2}x=\frac{1-{\sin }^{2}x{\cos }^{2}x}{2+{\sin }^{2}x{\cos }^{2}x}$

Answer 1

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

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

= RHS

Hence proved.