Question

Prove that ${\tan }^2\theta {\cos }^2\theta =1-{\cos }^2\theta$

Answer 1

$\begin{array}{c}L.H.S={\tan }^2\theta {\cos }^2\theta \\ =\left(\frac{1}{{\cos }^2\theta }-1\right){\cos }^2\theta \\ =\left(\frac{1}{{\cos }^2\theta }\right){\cos }^2\theta \\ =\left(\frac{{\cos }^2\theta }{{\cos }^2\theta }-{\cos }^2\theta \right)\\ =1-{\cos }^2\theta \\ =R.H.S\end{array}$