Q33 / 120 · NDA 2024 (Phase 2) · Mathematics
If $p \tan \left(\theta-30^{\circ}\right)=q \tan \left(\theta+120^{\circ}\right)$, then what is $\frac{(p+q)}{(p-q)}$
equal to? equal to?
- A$\sin 2 \theta$
- B$\cos 2 \theta$
- C$2 \sin 2 \theta$
- Correct answerD$2 \cos 2 \theta$
Solution
We have.
$\begin{aligned}
& p \tan \left(\theta-30^{\circ}\right)=q \tan \left(\theta+120^{\circ}\right) \\
& \Rightarrow \frac{p}{q}=\frac{\tan \left(\theta+120^{\circ}\right)}{\tan \left(\theta-30^{\circ}\right)} \\
& =\frac{\sin \left(\theta+120^{\circ}\right) \cos \left(\theta-30^{\circ}\right)}{\cos \left(\theta+120^{\circ}\right) \sin \left(\theta-30^{\circ}\right)}
\end{aligned}$
Using componendo and dividendo rule,
$\begin{gathered}
\Rightarrow \frac{p+q}{p-q}=\frac{\sin \left(\theta+120^{\circ}\right) \cos \left(\theta-30^{\circ}\right)+\cos \left(\theta+120^{\circ}\right) \sin \left(\theta-30^{\circ}\right)}{\sin \left(\theta+120^{\circ}\right) \cos \left(\theta-30^{\circ}\right)-\cos \left(\theta+120^{\circ}\right) \sin \left(\theta-30^{\circ}\right)} \\
\Rightarrow \frac{p+q}{p-q}=\frac{\sin \left[\left(\theta+120^{\circ}\right)+\left(\theta-30^{\circ}\right)\right]}{\sin \left[\left(\theta+120^{\circ}\right)-\left(\theta-30^{\circ}\right)\right]}=\frac{\sin \left(90^{\circ}+2 \theta\right)}{\sin \left(150^{\circ}\right)} \\
\Rightarrow \frac{p+q}{p-q}=\frac{\sin \left(90^{\circ}+2 \theta\right)}{\left(\frac{1}{2}\right)}=2 \cos 2 \theta
\end{gathered}$
Hence, the correct answer is option: 'D'.