Question

Solution of the equation $3\tan (\theta –15°)=\tan (\theta +15°)$ is

• A. $\theta =n\pi –\frac{\pi }{3}$
• B. $\theta =n\pi +\frac{\pi }{3}$
• C. $\theta =n\pi -\frac{\pi }{4}$
• D. $\theta =n\pi +\frac{\pi }{4}$

Answer 1

The correct option is D

$\theta =n\pi +\frac{\pi }{4}$

Explanation for the correct answer:

Determining the value of $\theta$.

We have the equation as:

$\Rightarrow 3\times \tan (\theta –15°)=\tan (\theta +15°)$

$\Rightarrow \frac{\tan (\theta +15°)}{\tan (\theta –15°)}=\frac{3}{1}$

By applying componendo and dividendo rule, i.e., $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}$

$$\begin{gathered} \Rightarrow \frac{3+1}{3-1}=\frac{\tan (\theta +15°)+\tan (\theta –15°)}{\tan (\theta +15°)–\tan (\theta –15°)} \\ \Rightarrow \frac{4}{2}=\frac{\frac{\sin (\theta +15°)}{\cos (\theta +15°)}+\frac{\sin (\theta –15°)}{\cos (\theta –15°)}}{\frac{\sin (\theta +15°)}{\cos (\theta +15°)}-\frac{\sin (\theta –15°)}{\cos (\theta –15°)}}\left[\because \tan x=\frac{\sin x}{\cos x}\right] \\ \Rightarrow 2=\frac{\sin (\theta +15°)\cos (\theta –15°)+\sin (\theta –15°)\cos (\theta +15°)}{\sin (\theta +15°)\cos (\theta –15°)–\sin (\theta –15°)\cos (\theta +15°)} \\ \Rightarrow 2=\frac{\sin (\theta +15°+\theta –15°)}{\sin (\theta +15°–\theta +15°)} \\ \Rightarrow 2=\frac{\sin 2\theta }{\sin 30°} \\ \Rightarrow 2\sin 30°=\sin 2\theta \\ \Rightarrow \sin 2\theta =2\left(\frac{1}{2}\right)\left[\because \sin 30°=\frac{1}{2}\right] \\ \Rightarrow \sin 2\theta =1 \\ \Rightarrow 2\theta ={\sin }^{-1}\left(1\right) \\ \Rightarrow 2\theta =2n\pi +\frac{\pi }{2}\left[\because {\sin }^{-1}\left(1\right)=2n\pi +\frac{\pi }{2}\right] \\ \Rightarrow \theta =n\pi +\frac{\pi }{4} \end{gathered}$$]

Therefore, the correct answer is option (D).