Question

If $\tan A=n\tan Band\sin A=m\sin B$, prove that ${\cos }^{2}A=\frac{m^2-1}{n^2-1}$.

Answer 1

$\sin A=m\sin B⟶\left(1\right)$

$\tan A=n\tan B$

$\frac{\sin A}{\cos A}=n\frac{\sin B}{\cos B}⟶\left(2\right)$

Substituting $\sin B$ from $equation$ $1,$ we get

$⟹\cos B=\frac{n}{m}\cos A⟶\left(3\right)$

${\sin }^{2}A=m^2{\sin }^{2}B$

$1-{\cos }^{2}A=m^2(1-{\cos }^{2}B)$

Substituting $equation$ $3,$ we get

$1-{\cos }^{2}A=m^2(1-\frac{n^2}{m^2}{\cos }^{2}A),$

${\cos }^{2}A=\frac{m^2-1}{n^2-1}$ $\left(proved\right)$