Question

If \[\dfrac{{(1 - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\] , then find the general value of \[\theta \] is
A) \[\left( 1 \right)\] \[n\pi \pm \dfrac{\pi }{6}\]
B) \[\left( 2 \right)\]\[n\pi + \dfrac{\pi }{6}\]
C) \[\left( 3 \right)\] \[2n\pi \pm \dfrac{\pi }{6}\]
D) \[\left( 4 \right)\] none of these

Answer 1

Hint: We have to find the general value of \[\theta \]. We solve this by using the trigonometric identities and the general values of the trigonometric functions . We also know the tan function is the ratio of sin function to cos function . Using the trigonometric identities of double angle and general solutions of trigonometric functions . On simplifying the equation we can find the value of \[\theta \].

Complete step-by-step answer:
All the trigonometric functions are classified into two categories or types as either sine function or cosine function . All the functions which lie in the category of sine functions are sin , cosec and tan functions on the other hand the functions which lie in the category of cosine functions are cos , sec and cot functions . The trigonometric functions are classified into these two categories on the basis of their property which is stated as : when the value of angle is substituted by the negative value of the angle then we get the negative value for the functions in the sine family and a positive value for the functions in the cosine family .
Given : \[\dfrac{{(1 - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta }} = \dfrac{1}{2} - - - - (1)\]
We know , \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Putting this in equation \[(1)\]
$\dfrac{{\left[ {1 - {{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}$
On simplifying , we get
\[\dfrac{{\left[ {\left( {\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\]
 We know , \[\cos \theta = \dfrac{1}{{\sec \theta }}\]
\[{\cos ^2}\theta - {\sin ^2}\theta = \dfrac{1}{2}\]
Also , \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
\[\cos 2\theta = \dfrac{1}{2}\]
We know , \[\cos \dfrac{\pi }{6} = \dfrac{1}{2}\]
\[\cos 2\theta = \cos \dfrac{\pi }{6}\]
Also we know ,
If \[\cos \theta = \cos \alpha \] , then
\[\theta = 2n\pi \pm \alpha \] , where \[n \in Z\]
Therefore , \[\theta = 2n\pi \pm \dfrac{\pi }{6}\]
Thus , the correct option is \[\left( 3 \right)\].
So, the correct answer is “ Option 3 ”.

Note: Equations involving trigonometric functions of a variable are called trigonometric equations . The solutions of a trigonometric equation for \[0 \leqslant x < 2\pi \] ( \[x\] is the angle of the trigonometric function ) are called principal solutions . The expression involving integer ’ \[n\] ‘ which gives all solutions of a trigonometric equation is called a general solution .
Various general formulas of trigonometric functions :
\[\sin \theta = \sin \alpha \] , then $\theta = n\pi \pm {( - 1)^n}\alpha$ , where \[n \in Z\]
\[\cos \theta = \cos \alpha \] , then \[\theta = 2n\pi \pm \alpha \] , where \[n \in Z\]
\[\tan \theta = \tan \alpha \] , then \[\theta = n\pi + \alpha \] , where \[n \in Z\]