Question

Show that $ta{n}^2\theta -si{n}^2\theta =ta{n}^2\theta si{n}^2\theta$

Answer 1

${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta \times {\sin }^2\theta$

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

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

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

$={\tan }^2\theta {\sin }^2\theta$ (proved)