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Question: If \[\dfrac{{(1 - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}\] , then find the gener...

If (1−tan⁡2θ)sec⁡2θ=12\dfrac{{(1 - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta }} = \dfrac{1}{2} , then find the general value of θ\theta is
A) (1)\left( 1 \right) nπ±π6n\pi \pm \dfrac{\pi }{6}
B) \left( 2 \right)$$$$n\pi + \dfrac{\pi }{6}
C) (3)\left( 3 \right) 2nπ±π62n\pi \pm \dfrac{\pi }{6}
D) (4)\left( 4 \right) none of these

Explanation

Solution

Hint : We have to find the general value of θ\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 θ\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 : (1−tan⁡2θ)sec⁡2θ=12−−−−(1)\dfrac{{(1 - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta }} = \dfrac{1}{2} - - - - (1)
We know , tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
Putting this in equation (1)(1)
[1−(sin⁡θcos⁡θ)2]sec⁡2θ=12\dfrac{{\left[ {1 - {{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}
On simplifying , we get
[(cos⁡2θ−sin⁡2θcos⁡2θ)]sec⁡2θ=12\dfrac{{\left[ {\left( {\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \right]}}{{{{\sec }^2}\theta }} = \dfrac{1}{2}
We know , cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }}
cos⁡2θ−sin⁡2θ=12{\cos ^2}\theta - {\sin ^2}\theta = \dfrac{1}{2}
Also , cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
cos⁡2θ=12\cos 2\theta = \dfrac{1}{2}
We know , cos⁡π6=12\cos \dfrac{\pi }{6} = \dfrac{1}{2}
cos⁡2θ=cos⁡π6\cos 2\theta = \cos \dfrac{\pi }{6}
Also we know ,
If cos⁡θ=cos⁡α\cos \theta = \cos \alpha , then
θ=2nπ±α\theta = 2n\pi \pm \alpha , where n∈Zn \in Z
Therefore , θ=2nπ±π6\theta = 2n\pi \pm \dfrac{\pi }{6}
Thus , the correct option is (3)\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⩽x<2π0 \leqslant x < 2\pi ( xx is the angle of the trigonometric function ) are called principal solutions . The expression involving integer ’ nn ‘ which gives all solutions of a trigonometric equation is called a general solution .
Various general formulas of trigonometric functions :
sin⁡θ=sin⁡α\sin \theta = \sin \alpha , then θ=nπ±(−1)nα\theta = n\pi \pm {( - 1)^n}\alpha , where n∈Zn \in Z
cos⁡θ=cos⁡α\cos \theta = \cos \alpha , then θ=2nπ±α\theta = 2n\pi \pm \alpha , where n∈Zn \in Z
tan⁡θ=tan⁡α\tan \theta = \tan \alpha , then θ=nπ+α\theta = n\pi + \alpha , where n∈Zn \in Z