
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
163.2k+ 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.
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