Question

Prove the following trigonometric identities:

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

Answer 1

Taking LHS we have

$ta{n}^2\theta –si{n}^2\theta$
$\to$ We know that $\frac{si{n}^2\theta }{co{s}^2\theta }=ta{n}^2\theta$ so substituting we have
$\to$ $\frac{si{n}^2\theta }{co{s}^2\theta }$- ${\sin }^2\theta$
$\to$ $\frac{si{n}^2\theta -si{n}^2\theta co{s}^2\theta }{co{s}^2\theta }$
$\to$ $\frac{si{n}^2\theta (1-co{s}^2\theta )}{co{s}^2\theta }$

$\to$ ${\sin }^2\theta ta{n}^2\theta$
$\to$ RHS
Hence proved.