Question

If $3(se{c}^2\theta +{\tan }^2\theta )=5$, then the general value of $\theta$ is

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

Answer 1

The correct option is C

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

Explanation for correct option:

Given, $3(se{c}^2\theta +{\tan }^2\theta )=5$

$\Rightarrow$ $\begin{array}{rcl}se{c}^2\theta +{\tan }^2\theta & =& \frac{5}{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{{\cos }^2\theta }+\frac{{\sin }^2\theta }{{\cos }^2\theta }& =& \frac{5}{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1+{\sin }^2\theta }{{\cos }^2\theta }& =& \frac{5}{3}\end{array}$

Using the identity $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{,}$

$\begin{array}{rcl}3(1+1–{\cos }^2\theta )& =& 5{\cos }^2\theta \end{array}$

$\Rightarrow$ $\begin{array}{rcl}6–3{\cos }^2\theta & =& 5{\cos }^2\theta \end{array}$

$\Rightarrow$$\begin{array}{rcl}5{\cos }^2\theta +3{\cos }^2\theta & =& 6\end{array}$

$\Rightarrow$ $\begin{array}{rcl}8{\cos }^2\theta & =& 6\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\cos }^2\theta & =& \frac{3}{4}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cos \theta & =& \pm \frac{\sqrt{3}}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cos \theta & =& \pm \cos \left(\frac{\pi }{6}\right)\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathbf{}\mathit{\theta }\mathbf{}& \mathbf{=}& \mathbf{}\mathit{n}\mathit{\pi }\mathbf{}\mathbf{\pm }\mathbf{}\frac{\mathbf{\pi }}{\mathbf{6}}\end{array}$

Hence, The correct answer is option (C).