Question

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

Answer 1

To prove $sin({90}^{\circ }-A)cos({90}^{\circ }-A)=\frac{tanA}{1+ta{n}^{2}A}$

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

$sin({90}^{\circ }-A)cos({90}^{\circ }-A)cosAsinA...\left(1\right)$

Consider $\frac{tanA}{1+ta{n}^{2}A}$

$\frac{tanA}{1+ta{n}^{2}A}=\frac{tanA}{se{c}^{2}A}$

$=\frac{SinA/cosA}{1/cos²A}$

$=\frac{SinA}{cosA}\times cos²A$

$\frac{tanA}{1+ta{n}^{2}A}$$=SinAcosA...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ , we get

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

Hence proved.