Question

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 1

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.