Question

Prove that ${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta .{\sin }^2\theta .Hence,deducethat{\tan }^2\theta \ge {\sin }^2\theta$

Answer 1

${\tan }^2\theta -{\sin }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta$

$=\frac{{\sin }^2\theta (1-{\cos }^2\theta )}{{\cos }^2\theta }$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }\cdot {\sin }^2\theta$

$\Rightarrow {\tan }^2{\sin }^2\theta -{\sin }^2\theta ={\tan }^2\theta \cdot {\sin }^2\theta$

$\Rightarrow {\tan }^2\theta {\sin }^2\theta \ge 0$ (always)

$\Rightarrow {\tan }^2\theta \ge {\sin }^2\theta$