
The most general value of$\theta $ satisfying the equations $\sin \theta =\sin \alpha $and $\cos \theta =\cos \alpha $is
A. \[2n\pi +\alpha \]
B. \[2n\pi -\alpha \]
C. \[n\pi +\alpha \]
D. \[n\pi -\alpha \]
Answer
162k+ views
Hint: To derive the general value of $\theta $, we will consider both the equations and divide them with each other so that equations becomes in terms of tan. Then we will apply the theorem which states that if $x$ and $y$are even multiples of $\dfrac{\pi }{2}$, then $\tan x=\tan y$ implies that $x=n\pi +y$, where $n\in Z$ and is an integer.
Complete step by step solution:We are given $\sin \theta =\sin \alpha $and $\cos \theta =\cos \alpha $we have to determine the general value of $\theta $.
First we will take both of the given equations $\sin \theta =\sin \alpha $and $\cos \theta =\cos \alpha $.
$\sin \theta =\sin \alpha .....(i)$
$\cos \theta =\cos \alpha ....(ii)$
Now we will divide the equation (i) by equation (ii).
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sin \alpha }{\cos \alpha }$
As we know that $\dfrac{\sin A}{\cos A}=\tan A$, so
$\tan \theta =\tan \alpha $
Applying the theorem here we will derive the general value of $\theta $.
$\begin{align}
& \tan \theta =\tan \alpha \\
& \theta =n\pi +\alpha
\end{align}$
The general value of $\theta $ is $\theta =n\pi +\alpha $ when .$\sin \theta +\cos \theta$
Option ‘C’ is correct
Note: We could have also derived the general value of $\theta $ without converting the equation in terms of tan.
We will now simplify the equation.
$\dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=0$
$\dfrac{\sin \theta \cos \alpha -\sin \alpha \cos \theta }{\cos \theta \cos \alpha }=0$
$\sin \theta \cos \alpha -\sin \alpha \cos \theta =0$
We will now use the formula $\sin (a-b)=\sin a\cos b-\cos a\sin b$.
$\begin{align}
& \sin (\theta -\alpha )=0 \\
& \sin (\theta -\alpha )=\sin 0 \\
\end{align}$
We know that the general solution of $\sin \theta =0$is $n\pi $ where $n\in Z$. So,
$\begin{align}
& \theta -\alpha =n\pi \\
& \theta =n\pi +\alpha \\
\end{align}$
As we have seen in the solutions above, we must remember all the theorems of general solutions of trigonometric functions of sin, cos and tan. With the help of general solutions of these functions we can easily derive the general solutions for sec, cosec and cot. As general solutions contains an integer $n$ such that $n\in Z$, we can get the principal solutions for the trigonometric functions by substituting the value of $n$.
Complete step by step solution:We are given $\sin \theta =\sin \alpha $and $\cos \theta =\cos \alpha $we have to determine the general value of $\theta $.
First we will take both of the given equations $\sin \theta =\sin \alpha $and $\cos \theta =\cos \alpha $.
$\sin \theta =\sin \alpha .....(i)$
$\cos \theta =\cos \alpha ....(ii)$
Now we will divide the equation (i) by equation (ii).
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sin \alpha }{\cos \alpha }$
As we know that $\dfrac{\sin A}{\cos A}=\tan A$, so
$\tan \theta =\tan \alpha $
Applying the theorem here we will derive the general value of $\theta $.
$\begin{align}
& \tan \theta =\tan \alpha \\
& \theta =n\pi +\alpha
\end{align}$
The general value of $\theta $ is $\theta =n\pi +\alpha $ when .$\sin \theta +\cos \theta$
Option ‘C’ is correct
Note: We could have also derived the general value of $\theta $ without converting the equation in terms of tan.
We will now simplify the equation.
$\dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \alpha }{\cos \alpha }=0$
$\dfrac{\sin \theta \cos \alpha -\sin \alpha \cos \theta }{\cos \theta \cos \alpha }=0$
$\sin \theta \cos \alpha -\sin \alpha \cos \theta =0$
We will now use the formula $\sin (a-b)=\sin a\cos b-\cos a\sin b$.
$\begin{align}
& \sin (\theta -\alpha )=0 \\
& \sin (\theta -\alpha )=\sin 0 \\
\end{align}$
We know that the general solution of $\sin \theta =0$is $n\pi $ where $n\in Z$. So,
$\begin{align}
& \theta -\alpha =n\pi \\
& \theta =n\pi +\alpha \\
\end{align}$
As we have seen in the solutions above, we must remember all the theorems of general solutions of trigonometric functions of sin, cos and tan. With the help of general solutions of these functions we can easily derive the general solutions for sec, cosec and cot. As general solutions contains an integer $n$ such that $n\in Z$, we can get the principal solutions for the trigonometric functions by substituting the value of $n$.
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