Question

I : lf $A+B+C=\pi$ and $\cos A=\cos B\cos C$, then $\cot B\cot C=\frac{1}{3}$.
II : If $5\sin B=\sin (2A+B)$, then $2\tan (A+B)=3\tan A$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

The correct option is B only II is true

Taking statement I

$\cos A=\cos B\cos C$

$\cos (\pi -(B+C\left)\right)=\cos B\cos C$

$-\cos (B+C)=\cos B\cos C$

$-\cos B\cos C+\sin B\sin C=\cos B\cos C$

$\sin B\sin C=2\cos B\cos C$

$\cot B\cot C=\frac{1}{2}$

Taking statement II

$5\sin B=\sin (2A+B)$

$\frac{\sin (2A+B)}{\sin B}=\frac{5}{1}$

Applying componendo and dividendo

$\frac{\sin (2A+B)+\sin B}{\sin (2A+B)-\sin B}=\frac{5+1}{5-1}$

$\frac{2\sin (A+B)\cos A}{2\cos (A+B)\sin A}=\frac{6}{4}=\frac{3}{2}$

$\frac{\tan (A+B)}{\tan A}=\frac{3}{2}$

$2\tan (A+B)=3\tan A$

Therefore, I is false and II is true.

Hence, option ${}^{\prime }{B}^{\prime }$ is correct.