
The general solution of \[{\sin ^2}\theta \sec \theta + \sqrt 3 \tan \theta =
0\] is
A. \[\theta = n\pi + {( - 1)^{n + 1}}\dfrac{\pi }{3},\theta = n\pi ,n \in z\]
B. \[\theta = {n^\pi }\], \[n \in z\]
C. \[\theta = n\pi + {( - )^{n + 1}}\dfrac{\pi }{3},n \in z\]
D. \[\theta = \dfrac{{n\pi }}{2},n \in Z\]
Answer
163.8k+ views
Hint: In order to answer this question, we will use the reciprocal relationships, by which \[\sec \theta \]will be equal to\[\dfrac{1}{{\cos \theta }}\], as well as the quotient relationship\[\tan \theta \], which is equal to\[\dfrac{{\sin \theta }}{{\cos \theta }}\]. By substituting these values into the given equation, we obtain an equation from which the common term \[\sin \theta \] is taken, and by moving all the other terms to the opposite side. As a result, the equation can have a general solution.
Formula Used: The trigonometric formulas of sin are:
\[{\sin ^2}\theta \]\[ = \dfrac{1}{{\cos \theta }}\]
\[\tan \theta \]\[ = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Complete step by step solution: We are given an equation with \[{\sin ^2}\theta \] and \[\tan \theta \], we have to substitute the values according to the quotient relation and reciprocal relation,
We will get the following equation:
\[{\sin ^2}\theta \sec \theta + \sqrt 3 \tan \theta = 0\]
Substitute \[\sec \theta \] is equal to\[\dfrac{1}{{\cos \theta }}\], in accordance with quotient relation, we will get,
\[{\sin ^2}\theta \dfrac{1}{{\cos \theta }} + \sqrt 3 \tan \theta = 0\]
Substitute \[\tan \theta \] is equal to \[\dfrac{{\sin \theta }}{{\cos \theta }}\], in accordance with reciprocal relation, we will get,
\[{\sin ^2}\theta \dfrac{1}{{\cos \theta }} + \sqrt 3 \dfrac{{\sin \theta }}{{\cos \theta }} = 0\]
Taking \[\dfrac{1}{{\cos \theta }}\]to the Right-hand side we will get,
\[\sin \theta \](\[\sin \theta \]+\[\sqrt 3 \])\[ = 0\]
Since, \[\sin \theta \] is a common term, we take 1 term from the above equation,
\[\sin \theta + \sqrt 3 = 0\]
Taking \[\sqrt 3 \]to the left side, positive \[\sqrt 3 \]will become negative \[\sqrt 3 \], as follows:
\[\sin \theta = - \sqrt 3 \]
\[\theta = n\pi \]
Option ‘B’ is correct
Note: To get the general solution for the given equation, \[{\sin ^2}\theta \sec \theta + \sqrt 3 \tan \theta = 0\]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 formulas of sin are:
\[{\sin ^2}\theta \]\[ = \dfrac{1}{{\cos \theta }}\]
\[\tan \theta \]\[ = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Complete step by step solution: We are given an equation with \[{\sin ^2}\theta \] and \[\tan \theta \], we have to substitute the values according to the quotient relation and reciprocal relation,
We will get the following equation:
\[{\sin ^2}\theta \sec \theta + \sqrt 3 \tan \theta = 0\]
Substitute \[\sec \theta \] is equal to\[\dfrac{1}{{\cos \theta }}\], in accordance with quotient relation, we will get,
\[{\sin ^2}\theta \dfrac{1}{{\cos \theta }} + \sqrt 3 \tan \theta = 0\]
Substitute \[\tan \theta \] is equal to \[\dfrac{{\sin \theta }}{{\cos \theta }}\], in accordance with reciprocal relation, we will get,
\[{\sin ^2}\theta \dfrac{1}{{\cos \theta }} + \sqrt 3 \dfrac{{\sin \theta }}{{\cos \theta }} = 0\]
Taking \[\dfrac{1}{{\cos \theta }}\]to the Right-hand side we will get,
\[\sin \theta \](\[\sin \theta \]+\[\sqrt 3 \])\[ = 0\]
Since, \[\sin \theta \] is a common term, we take 1 term from the above equation,
\[\sin \theta + \sqrt 3 = 0\]
Taking \[\sqrt 3 \]to the left side, positive \[\sqrt 3 \]will become negative \[\sqrt 3 \], as follows:
\[\sin \theta = - \sqrt 3 \]
\[\theta = n\pi \]
Option ‘B’ is correct
Note: To get the general solution for the given equation, \[{\sin ^2}\theta \sec \theta + \sqrt 3 \tan \theta = 0\]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
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Instantaneous Velocity - Formula based Examples for JEE

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

JEE Advanced 2025 Notes
