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

We have to find ${\cos }^{2}A$ in terms of $m$ and $n.$ This means that angle $B$ is to be eliminated from the given relations.

Now,

$\tan A=n\tan B$ $\Rightarrow$ $\tan B=$ $\frac{1}{n}$ $\tan A$ $\Rightarrow$ $\cot B=$ $\frac{n}{\tan A}$

and

$\sin A=m\sin B$ $\Rightarrow$ $\sin B$ = $\frac{1}{m}$ $\sin A$ $\Rightarrow$ $\text{cosec}B$ = $\frac{m}{\sin A}$

Substituting the values of $\cot B$ and $\text{cosec}B$ in ${\text{cosec}}^{2}B-{\cot }^{2}B=1$, we get,

$\Rightarrow \frac{m^2}{{\sin }^{2}A}-\frac{n^2}{{\tan }^{2}A}=1$

$\Rightarrow \frac{m^2}{{\sin }^{2}A}-\frac{n^2{\cos }^{2}A}{{\sin }^{2}A}=1$

$\Rightarrow \frac{m^2-n^2{\cos }^{2}A}{{\sin }^{2}A}=1$

$\Rightarrow m^2-n^2{\cos }^{2}A={\sin }^{2}A$

$\Rightarrow m^2-n^2{\cos }^{2}A=1-{\cos }^{2}A$

$\Rightarrow m^2-1=n^2{\cos }^{2}A-{\cos }^{2}A$

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

$\Rightarrow \frac{m^2-1}{n^2-1}={\cos }^{2}A$