Question

If $\sin \theta =\frac{m^2+2mn}{m^2+2mn+2n^2}$ prove that $\tan \theta =\frac{m^2+2mn}{2mn+2n^2}$.

Answer 1

$\sin \theta =\frac{m^2+2mn}{m^2+2mn+2n^2}$

Constant a right triangle

$ABC$ is a right triangle

$(m^2+2mn+2n^2{)}^2=(m^2+2mn{)}^2+x^2$

$(m^2+2mn+2n^2{)}^2-(m^2+2mn{)}^2+x^2<a^2-b^2=(a+b)(a-b)$

$(m^2+2mn+2n^2+m^2+2mn)(m^2+2mn+2n^2-m^2+2mn)=x^2$

$(2m^2+4mn+2a^2)\left(2n^2\right)=x^2$

$2\underset{}{\underset{}{(m^2+2mn+n^2)}}\left(2n^2\right)=x^2$

$2(m+n{)}^{2}2n^2=x^2$

$4n^2(m+n{)}^2=x^2$

$x=\sqrt{4n^2(m+n{)}^2}$

$=2n(m+n)$

$=2mn+2n^2$

$\tan \theta =\frac{m^2+2mn}{2mn+2n^2}$