Question

If $\frac{sin\,A}{sin\,B}=m$ and $\frac{cos\,A}{cos\,B}=n$ then find the value of $tan \,B$; $n^2 < 1 < m^2$.

• A. $n^2$
• B. $\pm \sqrt{\frac{1-n^{2}}{m^{2}-1}}$
• C. $n^{2}/\left(m^{2}-1\right)$
• D. $m^2$

Answer 1

Answer: (B)

$\pm \sqrt{\frac{1-n^{2}}{m^{2}-1}}$

Answer 2

Given, $\frac{sin\,A}{sin\,B}=m$ $\Rightarrow sin\,A=m\,sinB\quad\ldots\left(i\right)$ and $\frac{cos\,A}{cos\,B}=n$ $\Rightarrow cosA=ncosB\quad\ldots\left(ii\right)$ Squaring $\left(i\right)$ and $\left(ii\right)$ and then adding, we get $1=m^{2}\,sin^{2}\,B+n^{2}\,cos^{2}\,B$ $\Rightarrow \frac{1}{cos^{2}\,B}=m^{2}\, \frac{sin^{2}\,B}{cos^{2}\,B}+n^{2}$ (Dividing by $cos^2B$) $\Rightarrow sec^{2}B=m^{2}\,tan^{2}\,B+n^{2}$ $\Rightarrow 1+tan^{2}\,B=m^2\,tan^2\,B+n^{2}$ $\Rightarrow tan^{2}B=\frac{1-n^{2}}{m^{2}-1}$ $\Rightarrow tanB=\pm\sqrt{\frac{1-n^{2}}{m^{2}-1}}$.