Question

Prove $\to$ $\sin ({90}^o-\theta )\cos ({90}^o-\theta )=\frac{\tan \theta }{1+{\tan }^2\theta }$

Answer 1

Prove $sin(90-\theta ).cos(90-\theta )=\frac{tan\theta }{1+ta{n}^2\theta }$

$LHS=sin(90-\theta ).cos(90-\theta )$

$=cos\theta .sin\theta$

$=\frac{1}{2}.2sin\theta .cos\theta$

$=\frac{1}{2}sin2\theta$

$=\frac{1}{2}.\frac{2tan\theta }{1+ta{n}^2\theta }$

$=\frac{tan\theta }{1+ta{n}^2\theta }$

$=RHS$