Question

If $\sin A=\frac{2mn}{m^2+n^2}$, What is the value of tan A ?

• A. $\frac{2mn}{m^2+n^2}$
• B. $\frac{2mn}{m^2-n^2}$
• C. $\frac{m^2-n^2}{2mn}$
• D. $\frac{m^2+n^2}{m^2-n^2}$

Answer 1

The correct option is B $\frac{2mn}{m^2-n^2}$

$\sin A=\frac{2mn}{m^2+n^2}$
$\sin A=\frac{P}{H}=\frac{2mn}{m^2+n^2}$
Heren, $P=2mn$ and $H=m^2+n^2$
We know, $H^2=P^2+B^2$
$(m^2+n^2{)}^2=4m^2{n}^2+B^2$
$m^4+n^4+2m^2{n}^2=4m^2{n}^2+B^2$
$m^4+n^4-2m^2{n}^2=B^2$
$(m^2-n^2{)}^2=B^2$
$B=m^2-n^2$
Thus, $\tan A=\frac{P}{B}=\frac{2mn}{m^2-n^2}$