Question

If $cot\theta +cot\left(\frac{\pi }{4}+\theta \right)=2$, then the general value of $\theta$ is

• A. $2n\pi \pm \frac{\pi }{6}$
• B. $2n\pi \pm \frac{\pi }{3}$
• C. $n\pi \pm \frac{\pi }{3}$
• D. $n\pi \pm \frac{\pi }{6}$

Answer 1

The correct option is D

$n\pi \pm \frac{\pi }{6}$

Find the value of $\theta :$

Given, $cot\theta +cot\left(\frac{\pi }{4}+\theta \right)=2$

$\Rightarrow$ $cot\theta +\left[\frac{cot\left({\displaystyle \frac{\pi }{4}}\right)cot\theta –1}{cot\theta +cot\left(\frac{\pi }{4}\right)}\right]=2$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{t}\left(a+b\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{a}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{b}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{b}}\right]$

$\Rightarrow$ $cot\theta +\left[\frac{cot\theta –1}{cot\theta +1}\right]=2$

$\Rightarrow$ $cot\theta (cot\theta +1)+(cot\theta –1)=2(cot\theta +1)$

$\Rightarrow$ $co{t}^2\theta +cot\theta +cot\theta –1=2cot\theta +2$

$\Rightarrow$$co{t}^2\theta +2cot\theta –1–2cot\theta –2=0$

$\Rightarrow$ $co{t}^2\theta –3=0$

$\Rightarrow$ $co{t}^2\theta =3$

$\Rightarrow$ ${\tan }^2\theta =\frac{1}{3}$

$\Rightarrow$ ${\tan }^2\theta ={\tan }^2\left(\frac{\pi }{6}\right)$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\left(\frac{\pi }{6}\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}\right]$

$\mathbf{\therefore }\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathit{\pi }\mathbf{}\mathbf{\pm }\mathbf{}\frac{\mathbf{\pi }}{\mathbf{6}}$

Hence, Option ‘D’ is Correct.