Question

If $\frac{(1-\cos 2\theta )}{(1+\cos 2\theta )}=3,$ then the general value of $\theta$ is

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

Answer 1

The correct option is D

$n\pi \pm \frac{\pi }{3}$

The explanation for the correct option:

Step1. Calculate the value of the trigonometric ratio:

Given,$\frac{(1-\cos 2\theta )}{(1+\cos 2\theta )}=3,$

$$\begin{gathered} \Rightarrow \frac{\left[1-\left(1-2{\sin }^2\left(\theta \right)\right)\right]}{\left[1+\left(2{\cos }^2\left(\theta \right)-1\right)\right]}=3\mathit{[}\mathit{\because }\mathit{c}\mathit{o}\mathit{s}\mathit{\left(}\mathit{2}\mathit{\theta }\mathit{\right)}\mathit{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathit{2}}\mathit{\left(}\mathit{\theta }\mathit{\right)}\mathit{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathit{2}}\mathit{\left(}\mathit{\theta }\mathit{\right)}\mathit{=}\mathit{1}\mathit{-}\mathit{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathit{2}}\mathit{\left(}\mathit{\theta }\mathit{\right)}\mathit{=}\mathit{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathit{2}}\mathit{\left(}\mathit{\theta }\mathit{\right)}\mathit{-}\mathit{1}\mathit{]} \\ \Rightarrow \frac{1-1+2{\sin }^2\left(\theta \right)}{1+2{\cos }^2\left(\theta \right)-1}=3 \\ \Rightarrow \frac{2{\sin }^2\left(\theta \right)}{2{\cos }^2\left(\theta \right)}=3 \\ \Rightarrow \frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=3 \\ \Rightarrow {\tan }^2\left(\theta \right)=3\left[\mathbf{\because }\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\right] \\ \Rightarrow \tan \left(\theta \right)=\pm \sqrt{3} \end{gathered}$$

Step2. Find the general solution of $\theta$:

We know that

If $\tan \left(\mathrm{\theta }\right)=\tan \left(\mathrm{\alpha }\right)$ then the general solution of $\mathrm{\theta }=\mathrm{n\pi }+\mathrm{\alpha },\mathrm{n}\in \mathrm{Z}$

$$\begin{gathered} \tan \left(\theta \right)=\pm \sqrt{3} \\ \Rightarrow \tan \left(\theta \right)=\pm \tan \left(\frac{\mathrm{\pi }}{3}\right)\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{3}}\mathbf{]} \\ \Rightarrow \mathit{\theta }\mathbf{=}\mathit{n}\mathit{\pi }\mathbf{\pm }\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Hence, Option(D) is the correct answer.