Q63 / 160 · AP EAMCET 2023 (18 May Shift 1) · Mathematics
If $m \cdot \tan \left(\theta-30^{\circ}\right)=n \cdot \tan \left(\theta+120^{\circ}\right)$, then $\frac{m+n}{m-n}=$
- Correct answerA$2 \cos 2 \theta$
- B$2 \cos ^2 \theta$
- C$\tan 2 \theta$
- D$2 \sin 2 \theta$
Solution
Given : $\mathrm{m} \tan \left(\theta-30^{\circ}\right)=\mathrm{n} \tan (\theta+120)$
$\begin{aligned} & \Rightarrow \mathrm{m} \tan \left(\theta-30^{\circ}\right)=-\mathrm{n} \cot (\theta+30) \\ & \Rightarrow \frac{\mathrm{n}}{\mathrm{m}}=-\tan (\theta-30) \tan (\theta+30)\end{aligned}$
$\begin{aligned} & =-\left(\frac{\tan \theta-\tan 30}{1+\tan \theta \tan 30}\right)\left(\frac{\tan \theta \tan 30}{1-\tan \theta \tan 30}\right) \\ & =\frac{\tan ^2 30-\tan ^2 \theta}{1-\tan ^2 \theta \tan ^2 30}\end{aligned}$
Now, $\frac{m+n}{m-n}=\frac{1+\frac{n}{m}}{1-\frac{n}{m}}$
$=\frac{1+\frac{\tan ^2 30-\tan ^2 \theta}{1-\tan ^2 \theta \tan ^2 30}}{1-\frac{\tan ^2 30-\tan ^2 \theta}{1-\tan ^2 \theta \tan ^2 30}}$
$\begin{aligned} & =\frac{\left(1-\tan ^2 \theta\right)\left(1+\tan ^2 30\right)}{\left(1+\tan ^2 \theta\right)\left(1-\tan ^2 30\right)} \\ & =\frac{\left(\cos ^2 \theta-\sin ^2 \theta\right)\left(\cos ^2 30+\sin ^2 30\right)}{\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 30-\sin ^2 30\right)} \\ & =\frac{\cos 2 \theta \times 1}{1 \times \cos (2 \times 60)}=\frac{\cos 2 \theta}{\frac{1}{2}}\end{aligned}$
$=2 \cos 2 \theta$
Hence, the correct answer is option: 'A'.