Question

If $3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$, then the general value of $\theta $ is
A. $2n\pi + \dfrac{\pi }{6}$
B. $2n\pi \pm \dfrac{\pi }{6}$
C. $n\pi \pm \dfrac{\pi }{6}$
D. $n\pi \pm \dfrac{\pi }{3}$

Answer 1

Hint: Given, $3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$ . Firstly, we will convert the given equation in the form of $\sin \theta $ and $\cos \theta $ using the identity $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$. After simplifying we use ${\sin ^2}x + {\cos ^2}x = 1$ identity to convert the simplified equation into the form of $\cos \theta $. Then, we will solve the simplified equation to find the general value of $\theta $.

Complete step by step solution: 
 Given, $3({\tan ^2}\theta + {\sec ^2}\theta ) = 5$
We know $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$
$3\left( {\dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right) = 5$
$3\left( {\dfrac{{1 + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right) = 5$
Using identity ${\sin ^2}x = 1 - {\cos ^2}x$
$3\left( {\dfrac{{1 + 1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right) = 5$
$3\left( {\dfrac{{2 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right) = 5$
Simplifying
$6 - 3{\cos ^2}\theta = 5{\cos ^2}\theta $
$6 = 8{\cos ^2}\theta $
Dividing both sides with $8$
$\dfrac{6}{8} = {\cos ^2}\theta $
Taking square root on both sides
$\cos \theta = \pm \sqrt {\dfrac{3}{4}} $
$\cos \theta = \pm \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \theta = \pm \dfrac{\pi }{6}$
Hence, the general value of $\theta $ is $n\pi \pm \dfrac{\pi }{6}$
Therefore, option C is correct..

Note: Students should pay attention while using the identities to convert the equation if they do not they can make mistakes. After taking square root there will be two values one positive and other is negative most of the students avoid negative values which leads to incorrect solutions.