Question

Prove the following identities (1-17)

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

Answer 1

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



$$\begin{gathered} \mathrm{LHS}=\left(\frac{1}{se{c}^2\theta -{\cos }^2\theta }+\frac{1}{\cos e{c}^2\theta -{\sin }^2\theta }\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{1}{{\displaystyle \frac{1}{{\cos }^2\theta }}-{\cos }^2\theta }+\frac{1}{{\displaystyle \frac{1}{{\sin }^2\theta }}-{\sin }^2\theta }\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{{\cos }^2\theta }{1-{\cos }^4\theta }+\frac{{\sin }^2\theta }{1-{\sin }^4\theta }\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{{\cos }^2\theta \left(1-{\sin }^4\theta \right)+{\sin }^2\theta \left(1-{\cos }^4\theta \right)}{\left(1-{\cos }^4\theta \right)\left(1-{\sin }^4\theta \right)}\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{1-{\cos }^2\theta {\sin }^4\theta -{\cos }^4\theta {\sin }^2\theta }{\left(1+{\sin }^2\theta \right)\left(1-s{\mathrm{in}}^2\theta \right)\left(1+{\cos }^2\theta \right)\left(1-{\cos }^2\theta \right)}\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{1-{\cos }^2\theta {\sin }^4\theta -{\cos }^4\theta {\sin }^2\theta }{\left(1+{\sin }^2\theta \right).{\cos }^2\theta .\left(1+{\cos }^2\theta \right).{\sin }^2\theta }\right){\sin }^2\theta {\cos }^2\theta \\ =\left(\frac{1-{\cos }^2\theta {\sin }^4\theta -{\cos }^4\theta {\sin }^2\theta }{\left(1+{\sin }^2\theta \right)\left(1+{\cos }^2\theta \right)}\right) \\ =\frac{1-{\cos }^2\theta {\sin }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)}{2+{\sin }^2\theta .c{\mathrm{os}}^2\theta } \\ =\frac{1-{\cos }^2\theta {\sin }^2\theta }{2+{\sin }^2\theta .c{\mathrm{os}}^2\theta } \end{gathered}$$

= RHS

Hence proved.