Question

If $\frac{\sin A}{\sin B}=\frac{1}{3}$ and $\frac{\cos A}{\mathrm{cosB}}=2$ , then ${\cot }^{2}B$

• A. $\frac{32}{3}$
• B. $\frac{3}{32}$
• C. $\frac{8}{27}$
• D. $\frac{27}{8}$

Answer 1

Answer: (C)

$\frac{8}{27}$

$\frac{\sin A}{\sin B}=\frac{1}{3}$

$\frac{\cos A}{\cos B}=2$

$\Rightarrow \sin B=3\sin A$

$\Rightarrow {\sin }^{2}B=9{\sin }^{2}A$ (Squaring both the sides).

$\Rightarrow 1-{\cos }^{2}B=9(1-{\cos }^{2}A)$ $(\because {\sin }^{2}A+{\cos }^{2}A=1)$

$\Rightarrow 1-{\cos }^{2}B=9-9{\cos }^{2}A$

$\Rightarrow 9{\cos }^{2}A-{\cos }^{2}B=8$-------------(1)

Now $\frac{\cos A}{\cos B}\equiv 2$

$\Rightarrow \cos A=2\cos B$

Squaring both the sides we get,

$\therefore {\cos }^{2}A=4{\cos }^{2}B$---------------(2)

Substituting Equation (2) in (1), we get

$\Rightarrow 36{\cos }^{2}B-{\cos }^{2}B=8$

$\Rightarrow 35{\cos }^{2}B=8$

$\Rightarrow {\cos }^{2}B=\frac{8}{35}$ ${\sin }^{2}B=1-{\cos }^{2}B=\frac{27}{35}$

$\therefore {\cot }^{2}B=\frac{{\cos }^{2}B}{{\sin }^{2}B}=\frac{\frac{8}{35}}{\frac{27}{35}}=\frac{8}{27}$