Question

Solve
$\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{{\sin }^2\theta }{1}={\tan }^2\theta \cdot {\sin }^2\theta$

Answer 1

$\frac{si{n}^2\theta }{co{s}^2\theta }=si{n}^2\theta =ta{n}^2\theta si{n}^2\theta$

$si{n}^2\theta (\frac{1}{co{s}^2\theta }-1)=\frac{si{n}^2\theta \times si{n}^2\theta }{co{s}^2\theta }$

such that $si{n}^2\theta \ne 0$ & $co{s}^2\theta \ne 0$

we get , $1-co{s}^2\theta =si{n}^2\theta$ Hence proved.