Question

Prove that $\sin ({90}^0-A).\cos ({90}^0-A)=\frac{\tan A}{1+{\tan }^{2}A}$

Answer 1

$LHS=\sin ({90}^{\circ }-A)\cos ({90}^{\circ }-A)$

$=\sin A\cos A=\frac{1}{2}\sin 2A$

$=\frac{\tan A}{1+{\tan }^{2}A}=RHS$

$\because L.H.S=R.H.S$

$sin({90}^{\circ }-A)\cos ({90}^{\circ }-A)=\frac{\tan A}{1+{\tan }^{2}A}$

Hence Proved