Question

If $\frac{(1-{\tan }^2\mathrm{\theta })}{{\sec }^2\mathrm{\theta }}=\frac{1}{2},$ then the general value of $\theta$ is

• A. $\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{6}$
• B. $n\pi +\frac{\pi }{6}$
• C. $2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{6}$
• D. None of these

Answer 1

The correct option is A

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

Explanation of the correct option:

Step-1. Calculate the value of $\tan \left(\theta \right)$:

Given,$\frac{(1-{\tan }^2\mathrm{\theta })}{{\sec }^2\mathrm{\theta }}=\frac{1}{2},$

$$\begin{gathered} \Rightarrow \frac{(1-{\tan }^2\mathrm{\theta })}{(1+{\tan }^2\mathrm{\theta })}=\frac{1}{2}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{sec}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{\Rightarrow }\mathbf{}{\mathbf{sec}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}\mathbf{]} \\ \Rightarrow 2(1-{\tan }^2\mathrm{\theta })=1(1+{\tan }^2\mathrm{\theta }) \\ \Rightarrow 2-2t{\mathrm{an}}^2\mathrm{\theta }=1+{\tan }^2\mathrm{\theta } \\ \Rightarrow 2-1=2{\tan }^2\mathrm{\theta }+{\tan }^2\mathrm{\theta } \\ \Rightarrow 1=3{\tan }^2\mathrm{\theta } \\ \Rightarrow {\tan }^2\mathrm{\theta }=\frac{1}{3} \\ \Rightarrow \mathrm{tan\theta }=\pm \frac{1}{\sqrt{3}} \end{gathered}$$

Step2. Find the value of $\theta$:

We know that

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

We have $\mathrm{tan\theta }=\pm \frac{1}{\sqrt{3}}$

$$\begin{gathered} \Rightarrow \mathrm{tan\theta }=\pm \tan \left(\frac{\mathrm{\pi }}{6}\right)\left[\mathbf{\because }\mathbf{tan}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}\right] \\ \Rightarrow \mathit{\theta }\mathbf{=}\mathit{n}\mathbf{\pi }\mathbf{\pm }\frac{\mathbf{\pi }}{\mathbf{6}} \end{gathered}$$

Hence, Option(A) is the correct answer.