Question

Prove the following:
(iii) $se{c}^2\theta +\cos e{c}^2\theta =se{c}^2\theta \times \cos e{c}^2\theta$

Answer 1

$$\begin{gathered} \mathrm{LHS}=sec{}^2\theta +\cos e{c}^2\theta \\ =\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta \times {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta \times {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta }\times \frac{1}{{\sin }^2\theta } \\ =se{c}^2\theta \times \cos e{c}^2\theta \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$