Question

The most general values of $\theta$ satisfying $\tan \theta +\tan \left(\frac{3\mathrm{\pi }}{4}+\theta \right)=2$ are given by

• A. $\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}}$
• B. $\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}}$
• C. $\theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6},n\in \mathrm{\mathbb{Z}}$
• D. $\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6},n\in \mathbb{Z}$

Answer 1

The correct option is B

$\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}}$

Explanation for the correct option:

Given equation: $\tan \theta +\tan \left(\frac{3\pi }{4}+\theta \right)=2$

By simplifying the given equation, we get

$$\begin{gathered} \mathrm{tan\theta }+\tan \left(\frac{3\mathrm{\pi }}{4}+\mathrm{\theta }\right)=2 \\ \Rightarrow \mathrm{tan\theta }+\frac{\tan \left({\displaystyle \frac{3\mathrm{\pi }}{4}}\right)+\mathrm{tan\theta }}{1-\tan \left({\displaystyle \frac{3\mathrm{\pi }}{4}}\right)\mathrm{tan\theta }}=2\left[\because \tan (\mathrm{A}+\mathrm{B})=\frac{\mathrm{tanA}+\mathrm{tanB}}{1-\mathrm{tanA}\mathrm{tanB}}\right] \\ \Rightarrow \mathrm{tan\theta }+\frac{(-1)+\mathrm{tan\theta }}{1-(-1)\mathrm{tan\theta }}=2 \\ \Rightarrow \mathrm{tan\theta }+\frac{\mathrm{tan\theta }-1}{1+\mathrm{tan\theta }}=2 \\ \Rightarrow \frac{\mathrm{tan\theta }\left(1+\mathrm{tan\theta }\right)+\mathrm{tan\theta }-1}{1+\mathrm{tan\theta }}=2 \\ \Rightarrow \mathrm{tan\theta }\left(1+\mathrm{tan\theta }\right)+\mathrm{tan\theta }-1=2\left(1+\mathrm{tan\theta }\right) \\ \Rightarrow \mathrm{tan\theta }+{\tan }^2\mathrm{\theta }+\mathrm{tan\theta }-1=2+2\mathrm{tan\theta } \\ \Rightarrow {\tan }^2\mathrm{\theta }+2\mathrm{tan\theta }-1=2+2\mathrm{tan\theta } \\ \Rightarrow {\tan }^2\mathrm{\theta }-1=2 \\ \Rightarrow {\tan }^2\mathrm{\theta }=3 \\ \Rightarrow \mathrm{tan\theta }=\pm \sqrt{3} \\ \Rightarrow \mathrm{tan\theta }=\tan \left(\frac{\mathrm{\pi }}{3}\right)\mathrm{or}\tan \left(\frac{2\mathrm{\pi }}{3}\right) \\ \Rightarrow \mathrm{\theta }=\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{\mathbb{Z}} \end{gathered}$$

Therefore, the most general values of $\theta$ satisfying $\tan \theta +\tan \left(\frac{3\mathrm{\pi }}{4}+\theta \right)=2$ are given by $\theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}}$.

Hence, option B is correct .