
The solution of the equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$is
A. \[\theta =\dfrac{1}{2}\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right]\]
B. \[\theta =n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4}\]
C. \[\theta =\dfrac{n\pi }{2}+{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4}\]
D. None of these.
Answer
160.8k+ views
Hint: we will first convert the given equation in terms of sin and cos and simplify. Then in the resultant equation we derived we will use the formula of $\sin 2A$. After this we will square the equation on both sides of the equation and further simplify it We will then equate both the factors to zero and get a resultant equation in which we will apply the theorem which states that for all the real numbers $x,y$, $\sin x=\sin y$ implies that \[x=n\pi +{{(-1)}^{n}}y\] where \[n\in Z\].
Formula Used: $\sin 2A=2\sin A\cos A$
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Complete step by step solution: We are given an equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$ and we have to determine the general solution of this equation.
We will first take the given equation and convert it in terms of sin and cos using formula $\sec \theta =\dfrac{1}{\cos \theta }$ and $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and simplify it.
$\begin{align}
& \sec \theta -\cos ec\theta =\dfrac{4}{3} \\
& \dfrac{1}{\cos \theta }-\dfrac{1}{\sin \theta }=\dfrac{4}{3} \\
& \dfrac{\sin \theta -\cos \theta }{\sin \theta \cos \theta }=\dfrac{4}{3} \\
& 3\left( \sin \theta -\cos \theta \right)=4\sin \theta \cos \theta \\
& 3\left( \sin \theta -\cos \theta \right)=2\left( 2\sin \theta \cos \theta \right)
\end{align}$
We will now use the double angle formula of sin.
$3\left( \sin \theta -\cos \theta \right)=2\sin 2\theta $
Squaring on both sides of the equation,
$\begin{align}
& 9{{\left( \sin \theta -\cos \theta \right)}^{2}}=4{{\sin }^{2}}2\theta \\
& 9({{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta )=4{{\sin }^{2}}2\theta \\
\end{align}$
As we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we will substitute this and formula of $\sin 2A$use in the above equation.
$9(1-\sin 2\theta )=4{{\sin }^{2}}2\theta $
Now we will simplify the equation to form a quadratic equation.
$4{{\sin }^{2}}2\theta +9\sin 2\theta -9=0$
We will now factorize the equation.
$\begin{align}
& 4{{\sin }^{2}}2\theta +9\sin 2\theta -9=0 \\
& 4{{\sin }^{2}}2\theta +12\sin 2\theta -3\sin 2\theta -9=0 \\
& 4\sin 2\theta (\sin 2\theta +3)-3(\sin 2\theta +3)=0 \\
& (\sin 2\theta +3)(4\sin 2\theta -3)=0
\end{align}$
We will now equate both the factors to zero.
$\begin{align}
& (\sin 2\theta +3)=0 \\
& \sin 2\theta =-3
\end{align}$ or $\begin{align}
& (4\sin 2\theta -3)=0 \\
& 4\sin 2\theta =3 \\
& \sin 2\theta =\dfrac{3}{4}
\end{align}$
As we know that the value of sine lies in the interval of $\left[ -1,1 \right]$ but in $\sin 2\theta =-3$ , $-3<-1$ so this value will be discarded.
Now,
We can write $\dfrac{3}{4}$ in terms of sin as $\sin \left( {{\sin }^{-1}}\dfrac{3}{4} \right)$. So,
$\sin 2\theta =\sin \left( {{\sin }^{-1}}\dfrac{3}{4} \right)$
Applying the theorem here we will get the general solution of the equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$.
\[\begin{align}
& 2\theta =\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right] \\
& \theta =\dfrac{1}{2}\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right]
\end{align}\]
The solution of the equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$is \[\theta =\dfrac{1}{2}\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right]\]
Option ‘A’ is correct
Note: There are theorems for the general solution of function sin, cos, and tan only that is why we have to convert the given equation in terms of sin and cos.
Formula Used: $\sin 2A=2\sin A\cos A$
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Complete step by step solution: We are given an equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$ and we have to determine the general solution of this equation.
We will first take the given equation and convert it in terms of sin and cos using formula $\sec \theta =\dfrac{1}{\cos \theta }$ and $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and simplify it.
$\begin{align}
& \sec \theta -\cos ec\theta =\dfrac{4}{3} \\
& \dfrac{1}{\cos \theta }-\dfrac{1}{\sin \theta }=\dfrac{4}{3} \\
& \dfrac{\sin \theta -\cos \theta }{\sin \theta \cos \theta }=\dfrac{4}{3} \\
& 3\left( \sin \theta -\cos \theta \right)=4\sin \theta \cos \theta \\
& 3\left( \sin \theta -\cos \theta \right)=2\left( 2\sin \theta \cos \theta \right)
\end{align}$
We will now use the double angle formula of sin.
$3\left( \sin \theta -\cos \theta \right)=2\sin 2\theta $
Squaring on both sides of the equation,
$\begin{align}
& 9{{\left( \sin \theta -\cos \theta \right)}^{2}}=4{{\sin }^{2}}2\theta \\
& 9({{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta )=4{{\sin }^{2}}2\theta \\
\end{align}$
As we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we will substitute this and formula of $\sin 2A$use in the above equation.
$9(1-\sin 2\theta )=4{{\sin }^{2}}2\theta $
Now we will simplify the equation to form a quadratic equation.
$4{{\sin }^{2}}2\theta +9\sin 2\theta -9=0$
We will now factorize the equation.
$\begin{align}
& 4{{\sin }^{2}}2\theta +9\sin 2\theta -9=0 \\
& 4{{\sin }^{2}}2\theta +12\sin 2\theta -3\sin 2\theta -9=0 \\
& 4\sin 2\theta (\sin 2\theta +3)-3(\sin 2\theta +3)=0 \\
& (\sin 2\theta +3)(4\sin 2\theta -3)=0
\end{align}$
We will now equate both the factors to zero.
$\begin{align}
& (\sin 2\theta +3)=0 \\
& \sin 2\theta =-3
\end{align}$ or $\begin{align}
& (4\sin 2\theta -3)=0 \\
& 4\sin 2\theta =3 \\
& \sin 2\theta =\dfrac{3}{4}
\end{align}$
As we know that the value of sine lies in the interval of $\left[ -1,1 \right]$ but in $\sin 2\theta =-3$ , $-3<-1$ so this value will be discarded.
Now,
We can write $\dfrac{3}{4}$ in terms of sin as $\sin \left( {{\sin }^{-1}}\dfrac{3}{4} \right)$. So,
$\sin 2\theta =\sin \left( {{\sin }^{-1}}\dfrac{3}{4} \right)$
Applying the theorem here we will get the general solution of the equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$.
\[\begin{align}
& 2\theta =\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right] \\
& \theta =\dfrac{1}{2}\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right]
\end{align}\]
The solution of the equation $\sec \theta -\cos ec\theta =\dfrac{4}{3}$is \[\theta =\dfrac{1}{2}\left[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}\dfrac{3}{4} \right]\]
Option ‘A’ is correct
Note: There are theorems for the general solution of function sin, cos, and tan only that is why we have to convert the given equation in terms of sin and cos.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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 In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
