Question

If $2\tan A=3\tan B$, then $\frac{\sin 2B}{5-\cos 2B}$ is equal to

• A. $\tan A–\tan B$
• B. $\tan (A–B)$
• C. $\tan (A+B)$
• D. $\tan (A+2B)$

Answer 1

The correct option is B

$\tan (A–B)$

Explanation for the correct option:

Step 1. Find the value of $\frac{\sin 2B}{5-\cos 2B}$:

Given, $2\tan A=3\tan B$ ….(1)

$\frac{\sin 2B}{5-\cos 2B}=\frac{{\displaystyle \frac{2\tan B}{1+{\tan }^{2}B}}}{5–\left({\displaystyle \frac{1–{\tan }^{2}B}{1+{\tan }^{2}B}}\right)}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{t}\mathbf{a}\mathbf{n}\left({\displaystyle \frac{\theta }{2}}\right)}{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\left({\displaystyle \frac{\theta }{2}}\right)}\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\left({\displaystyle \frac{\theta }{2}}\right)}{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\left({\displaystyle \frac{\theta }{2}}\right)}\right]$

$$\begin{gathered} =\frac{2\tan B}{\left[5(1+{\tan }^{2}B)–1+{\tan }^{2}B\right]} \\ =\frac{2\tan B}{\left[5+5{\tan }^{2}B–1+{\tan }^{2}B\right]} \\ =\frac{2\tan B}{\left[4+6{\tan }^{2}B\right]} \\ =\frac{2\tan B}{\left[2\left(2+3{\tan }^{2}B\right)\right]} \\ =\frac{\tan B}{\left(2+3\tan B\tan B\right)} \end{gathered}$$

Step 2. Divide the numerator and denominator by 2:

$\Rightarrow \frac{\left({\displaystyle \frac{1}{2}}\right)\times \tan B}{\left[1+\left({\displaystyle \frac{3}{2}}\right)\times \tan B\times \tan B\right]}$

$\Rightarrow \frac{\left[\left({\displaystyle \frac{3}{2}}\right)\times \tan B–\tan B\right]}{\left[1+\tan A\tan B\right]}$ [from equation(1)]

$\Rightarrow \frac{t\mathrm{an}A–\tan B}{1+\tan A\tan B}$

$\Rightarrow \mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathit{A}\mathbf{}\mathbf{–}\mathbf{}\mathit{B}\mathbf{)}$

Hence, Option ‘B’ is Correct.