Question

If $\frac{sinA}{sinB}=m$ and $\frac{cosA}{cosB}=n$ then find the value of $tanB$;
$n^2<1<m^2$.

• 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)

Given, $\frac{sinA}{sinB}=m$
$\Rightarrow sinA=msinB\dots \left(i\right)$
and $\frac{cosA}{cosB}=n$
$\Rightarrow cosA=ncosB\dots \left(ii\right)$
Squaring $\left(i\right)$ and $\left(ii\right)$ and then adding, we get
$1=m^{2}si{n}^{2}B+n^{2}co{s}^{2}B$
$\Rightarrow \frac{1}{co{s}^{2}B}=m^2\frac{si{n}^{2}B}{co{s}^{2}B}+n^2$ (Dividing by $co{s}^{2}B$)
$\Rightarrow se{c}^{2}B=m^{2}ta{n}^{2}B+n^2$
$\Rightarrow 1+ta{n}^{2}B=m^{2}ta{n}^{2}B+n^2$
$\Rightarrow ta{n}^{2}B=\frac{1-n^2}{m^2-1}$
$\Rightarrow tanB=\pm \sqrt{\frac{1-n^2}{m^2-1}}$.