Question

If $m\tan \left(\theta -30°\right)=n\tan \left(\theta +120°\right)$, then $\frac{m+n}{m-n}$ is equal to

• A. $2\cos \left(2\theta \right)$
• B. $\cos \left(2\theta \right)$
• C. $2\sin \left(2\theta \right)$
• D. $\sin \left(2\theta \right)$

Answer 1

The correct option is A

$2\cos \left(2\theta \right)$

Explanation for the correct option:

It is given that, $m\tan \left(\theta -30°\right)=n\tan \left(\theta +120°\right)$

$$\begin{gathered} \Rightarrow \frac{m}{n}=\frac{\tan \left(\theta +120°\right)}{\tan \left(\theta -30°\right)} \\ \Rightarrow \frac{m+n}{m-n}=\frac{\tan \left(\theta +120°\right)+\tan \left(\theta -30°\right)}{\tan \left(\theta +120°\right)-\tan \left(\theta -30°\right)} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \frac{\sin \left(\theta +120°\right)}{\cos \left(\theta +120°\right)}}+{\displaystyle \frac{\sin \left(\theta -30°\right)}{\cos \left(\theta -30°\right)}}}{{\displaystyle \frac{\sin \left(\theta +120°\right)}{\cos \left(\theta +120°\right)}}-{\displaystyle \frac{\sin \left(\theta -30°\right)}{\cos \left(\theta -30°\right)}}} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \frac{\sin \left(\theta +120°\right)\cos \left(\theta -30°\right)+\sin \left(\theta -30°\right)\cos \left(\theta +120°\right)}{\cos \left(\theta +120°\right)\cos \left(\theta -30°\right)}}}{{\displaystyle \frac{\sin \left(\theta +120°\right)\cos \left(\theta -30°\right)-\sin \left(\theta -30°\right)\cos \left(\theta +120°\right)}{\cos \left(\theta +120°\right)\cos \left(\theta -30°\right)}}} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \sin \left(\theta +120°\right)\cos \left(\theta -30°\right)+\sin \left(\theta -30°\right)\cos \left(\theta +120°\right)}}{{\displaystyle \sin \left(\theta +120°\right)\cos \left(\theta -30°\right)-\sin \left(\theta -30°\right)\cos \left(\theta +120°\right)}} \end{gathered}$$

It is known that, $\sin \left(A+B\right)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)$ and $\sin \left(A-B\right)=\sin \left(A\right)\cos \left(B\right)-\cos \left(A\right)\sin \left(B\right)$.

Thus, $\frac{m+n}{m-n}=\frac{{\displaystyle \sin \left[\left(\theta +120°\right)+\left(\theta -30°\right)\right]}}{{\displaystyle \sin \left[\left(\theta +120°\right)-\left(\theta -30°\right)\right]}}$

$$\begin{gathered} \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \sin \left(2\theta +90°\right)}}{{\displaystyle \sin \left(150°\right)}} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \sin \left(2\theta +90°\right)}}{{\displaystyle \sin \left(60°+90°\right)}} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \sin \left(2\theta +90°\right)}}{{\displaystyle \sin \left(60°+90°\right)}}\left[\because \sin \left(\alpha +90°\right)=\cos \left(\alpha \right)\right] \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \cos \left(2\theta \right)}}{{\displaystyle \cos \left(60°\right)}} \\ \Rightarrow \frac{m+n}{m-n}=\frac{{\displaystyle \cos \left(2\theta \right)}}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{m+n}{m-n}=2\cos \left(2\theta \right) \end{gathered}$$

Therefore, $\frac{m+n}{m-n}$ is equal to $2\cos \left(2\theta \right)$.

Hence, (A) is the correct option.