Question

In triangle $ABC$, if $\sin A\cos B=\frac{1}{4}$ and $3\tan A=\tan B$, then ${\cot }^{2}A$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

$3\tan A=\tan B$ ...(1)
$\Rightarrow \frac{3\cos B}{\sin B}=\frac{\cos A}{\sin A}$
$\Rightarrow 3\left(\frac{a^2+c^2-b^2}{2ac}\right)\frac{1}{b}=\left(\frac{b^2+c^2-a^2}{2bc}\right)\frac{1}{a}$ ..{ Using Sine and Cosine rule}
$\Rightarrow a^2+\frac{c^2}{2}=b^2$ ...(2)

Given that: $\sin A\cos B=\frac{1}{4}$

$\Rightarrow \sin A\left(\frac{a^2+c^2-b^2}{2ac}\right)=\frac{1}{4}$ ...{ cosine rule }

$\Rightarrow \sin A\left(\frac{c^2-\frac{c^2}{2}}{2ac}\right)=\frac{1}{4}$ ...{ From (2)}

$\Rightarrow \frac{a}{\sin A}=\frac{c}{1}=\frac{c}{\sin \frac{\pi }{2}}$

$\therefore C=\frac{\pi }{2}\{Sinerule:\frac{a}{\sin A}=\frac{c}{\sin C}\}$

Given that: $3\tan A=\tan B$
$\Rightarrow 3\tan A=\tan (\pi -A-C)\Rightarrow 3\tan A=\tan (\pi -A-\frac{\pi }{2})\Rightarrow 3\tan A=\cot A\therefore {\cot }^{2}A=3$
Ans: B