Question

If $m=\frac{\sin A}{\sin B},n=\frac{\cos A}{\cos B}$, then prove that $(m^2-n^2){\sin }^{2}B=1-n^2$

Answer 1

$(m^2-n^2)$

$={\left(\frac{\sin A}{\sin B}\right)}^2-{\left(\frac{\cos A}{\cos B}\right)}^2$

$=\left(\frac{{\sin }^{2}A}{{\sin }^{2}B}\right)-\left(\frac{{\cos }^{2}A}{{\cos }^{2}B}\right)$

$\Rightarrow (m^2-n^2){\sin }^{2}B=({\sin }^{2}A-{\cos }^{2}A{\tan }^{2}B)$

$=1-{\cos }^{2}A-{\cos }^{2}A{\tan }^{2}B$

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

$=1-{\cos }^{2}A{\sec }^{2}B$

$=1-\frac{{\cos }^{2}A}{{\cos }^{2}B}$

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

$=1-n^2$