Question

If $2\tan A=3\tan B$, find the value of $\frac{\sin 2B}{5-\cos 2B}$

• A. $\sin (A-B)$
• B. $\cos (A-B)$
• C. $\cot (A-B)$
• D. $\tan (A-B)$

Answer 1

The correct option is D $\tan (A-B)$

$2\tan A=3\tan B$

$\Rightarrow \tan A=\frac{3}{2}\tan B$

$\frac{\sin 2B}{5-\cos 2B}=\frac{\sin 2B}{4+(1-\cos 2B)}$

$=\frac{\sin 2B}{4+2{\sin }^{2}B}$

$=\frac{\sin B\cos B}{2+si{n}^{2}B}$

$=\frac{\frac{\sin B}{\cos B}.\cos 2B}{2{\cos }^{2}B+3{\sin }^{2}B}$

$=\frac{\tan B.{\cos }^{2}B}{2{\cos }^{2}B+3{\sin }^{2}B}$

$=\frac{\tan B}{2+3{\tan }^{2}B}$

$=\frac{3/2\tan B-\tan B}{1+\frac{3}{2}{\tan }^{2}B}$

$=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$=\tan (A-B)$