
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
162.3k+ views
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.
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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
