Question : If $\tan A = n\tan B$ and $\sin A=m \sin B$, then the value of $\cos^{2}A$ is:
Option 1: $\frac{m^{2}-1}{n^{2}+1}$
Option 2: $\frac{m^{2}+1}{n^{2}-1}$
Option 3: $\frac{m^{2}+1}{n^{2}+1}$
Option 4: $\frac{m^{2}-1}{n^{2}-1}$
Correct Answer: $\frac{m^{2}-1}{n^{2}-1}$
Solution :
$\sin A = m\sin B$--------------(i)
$\tan A = n\tan B$
$\frac{\sin A}{\cos A}$ = $n\frac{\sin B}{\cos B}$------------------(ii)
Substituting $\sin B$ from equation (i),
⇒ $\cos B$ = $\frac{n}{m}$$\cos A$----------(iii)
Squaring both sides of equation (i),
⇒ $\sin^{2} A$ = $m^{2}\sin^{2}B$
⇒ 1–$\cos^{2} A$ = $m^{2}(1–\cos^{2} B)$
Substituting equation (iii),
1–$\cos^{2} A$ = $m^{2}(1–\frac{n^{2}}{m^{2}}$$\cos^{2} A)$
$\cos^{2} A$ = $\frac{m^{2}-1}{n^{2}-1}$
Hence, the correct answer is $\frac{m^{2}-1}{n^{2}-1}$.
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