Question

General value of \[\theta \] satisfying the equation \[{\tan ^2}\theta + \sec 2\theta= 1\] is
A. m\[\pi \], n\[\pi \]\[ + \dfrac{\pi }{3}\]
B. m\[\pi \], n\[\pi \] \[ \pm \dfrac{\pi }{3}\]
C. m\[\pi \], n\[\pi \] \[ \pm \dfrac{\pi }{6}\]
D. None of these (Where m and n are integers)

Answer 1

Hint: We can write the given equation in terms of \[\tan \theta \] and by equating it to 1 on the right-hand side, we will take the LCM on the left-hand side and simplify the expression that is obtained by reaching to the other side of the trigonometrical identity. left-hand side of the given equation. We will substitute the trigonometrical relation for \[\sec 2\theta \] that is \[\dfrac{1}{{\cos 2\theta }}\]which is equivalent to the square relation, \[\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\].

Formula Used:
The trigonometric relations that are used are:
\[\sec 2\theta \]=\[\dfrac{1}{{\cos 2\theta }}\]
\[\dfrac{1}{{\cos 2\theta }}\]=\[\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\]

Complete step by step solution:
 We are given a trigonometrical equation for which the general value of \[\theta \] has to be found.
\[{\tan ^2}\theta + \sec 2\theta = 1\]
 To do so, we will use the trigonometrical relation in the equation by substituting the value of, as follows:
\[\sec 2\theta \]=\[\dfrac{1}{{\cos 2\theta }}\]
We also know that the trigonometrical relation of \[\dfrac{1}{{\cos 2\theta }}\] is equivalent to tangent terms,
\[\dfrac{1}{{\cos 2\theta }}\]=\[\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\]
The equation can be written as,
\[{\tan ^2}\theta \]\[ + \]\[\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\]\[ = 1\]
By simplifying the terms on the left-hand side by taking L.C.M, we will get the following,
\[{\tan ^2}\theta \]\[(1 - {\tan ^2}\theta )\]\[ + \]\[1 + {\tan ^2}\theta \]\[ = 1 - {\tan ^2}\theta \]
By multiplying \[{\tan ^2}\theta \] with \[(1 - {\tan ^2}\theta )\], we will get
\[{\tan ^2}\theta \]\[ - {\tan ^4}\theta \]\[ + \]\[1 + {\tan ^2}\theta \]\[ - 1 + {\tan ^2}\theta \]\[ = 0\]
By simplifying the above derived trigonometrical identities, we will get the following,
\[3{\tan ^2}\theta \]\[ - {\tan ^4}\theta \]\[ = 0\]
Now, take the common term outside, which can be written as,
\[{\tan ^2}\theta \]\[(3 - {\tan ^2}\theta )\]\[ = 0\]
From the above trigonometrical identity, we can understand that,
\[\tan \theta = 0\] or
It will be equal to,
\[\tan \theta = \pm \sqrt 3 \]
If \[\tan \theta = 0\], then \[\theta \]=m\[\pi \], where m is an integer and if \[\tan \theta = \pm \sqrt 3 \], then \[\theta \]=n\[\pi \pm \dfrac{\pi }{3}\] where n is an integer.
       Thus, \[\theta \]=m\[\pi \], n\[\pi \pm \dfrac{\pi }{3}\]is the value satisfying the given equation.

Option ‘B’ is correct

Note:
To get the general solution for the given equation, \[{\tan ^2}\theta + \sec 2\theta = 1\]we have to use trigonometry identities that are equal to their reciprocal relations and to their quotient relations in accordance with the required solution. The angle \[\theta \] can be generated from the given equation by substituting it in the given equation. In order to find the solution for the given trigonometrical identity we should start solving from the left-hand side or right-hand side and by applying trigonometrical relations we should come to the other side.