Question

If \[\dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\], then the general value of \[\theta \] is
A. \[n\pi \pm \dfrac{\pi }{6}\]
B. \[n\pi + \dfrac{\pi }{6}\]
C. \[2n\pi \pm \dfrac{\pi }{6}\]
D. None of these

Answer 1

Hint: In the above question, we need to determine the general value of \[\theta \]. For that, we will first simplify a given trigonometric equation by using trigonometric identities and then use the general value of a particular trigonometric function to find the required answer.

Formula used:
We have been using the following formulas:
1. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2. \[\cos \theta = \dfrac{1}{{\sec \theta }}\]
3. \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]

Complete step-by-step solution:
Given that \[\dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}...\left( 1 \right)\]
Now we know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Now we substitute this formula in equation (1), and we get
\[\dfrac{{\left[ {1 - {{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\]
Now we simplify the above equation, we get
\[
  \dfrac{{\left[ {1 - \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2} \\
  \dfrac{{\left[ {\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2} \\
  \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta \times {{\sec }^2}\theta }} = \dfrac{1}{2} \\
 \]
Now we know that \[\cos \theta = \dfrac{1}{{\sec \theta }}\]
Now by applying this formula in our equation, we get
\[\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta \times \dfrac{1}{{{{\cos }^2}\theta }}}} = \dfrac{1}{2}\]
By cancelling like terms, we get
\[{\cos ^2}\theta - {\sin ^2}\theta = \dfrac{1}{2}\]
We know that \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
So, \[\cos \,2\theta = \dfrac{1}{2}\]
Now we know that \[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\]
By applying the above value, we get
\[\cos \,2\theta = \cos \dfrac{\pi }{3}\]
Now we know that if \[\cos \theta = \cos \alpha \] then its general solution is \[\theta = 2n\pi \pm \alpha \]
Therefore, then general solution of \[\theta \] is \[2n\pi \pm \dfrac{\pi }{6}\]
Hence, option (C) is the correct option.

Note: There is an alternate method to solve the given question shown below:
Given that \[\dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}...\left( 1 \right)\]
We know that \[{\sec ^2}A = 1 + {\tan ^2}A\]
Now we apply this formula in equation (1), we get
\[\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \dfrac{1}{2}\]
By cross multiplying the terms and simplifying, we get
\[
  2\left( {1 - {{\tan }^2}\theta } \right) = 1 + {\tan ^2}\theta \\
  2 - 2{\tan ^2}\theta = 1 + {\tan ^2}\theta \\
  2 - 1 = {\tan ^2}\theta + 2{\tan ^2}\theta \\
  1 = {\tan ^2}\theta + 2{\tan ^2}\theta
 \]
Further simplifying, we get
\[
  3{\tan ^2}\theta = 1 \\
  {\tan ^2}\theta = \dfrac{1}{3} \\
  \tan \theta = \pm \dfrac{1}{{\sqrt 3 }}
 \]
We know that \[\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}\]
So, \[\tan \theta = \tan \dfrac{\pi }{6}\]
Now we know that if \[\tan \theta = \tan \alpha \] then the general solution is \[\theta = n\pi \pm \alpha \]
\[\therefore \theta = n\pi \pm \dfrac{\pi }{6}\]
Therefore, if \[\dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\], then the general value of \[\theta \] is \[n\pi \pm \dfrac{\pi }{6}\] ,where n is integer.
Hence, option (A) is correct