Question

Find the value of $\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }$
A. $\dfrac{1}{2}\tan \theta $
B. $\dfrac{1}{2}\cot \theta $
C. $\tan \theta $
D. $\cot \theta $

Answer 1

Hint: At first, we find the value of $\sin 2\theta $ in terms of $\sin \theta $ and $\cos \theta $ . We then find the value of $\cos 2\theta $ in terms of $\cos \theta $ alone. Then, we take some terms common from the numerator and the denominator. After cancelling out the terms, we get the required answer.

Complete step by step answer:
The trigonometric expression that we are given in this problem is,
$\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }$
We know the trigonometric identities which say that,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
Now, if we put the values of A and B both equal to $\theta $ , then, we will get,
$\begin{align}
  & \sin \left( \theta +\theta \right)=\sin \theta \cos \theta +\cos \theta \sin \theta \\
 & \Rightarrow \sin 2\theta =2\sin \theta \cos \theta ....\left( i \right) \\
\end{align}$
Doing the same for $\cos \left( A+B \right)$ , we will get,
$\begin{align}
  & \cos \left( \theta +\theta \right)=\cos \theta \cos \theta -\sin \theta \sin \theta \\
 & \Rightarrow \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
\end{align}$
We also know the trigonometric identity which states that,
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Using the above trigonometric identity in the equation for $\cos 2\theta $ , we will get,
$\begin{align}
  & \Rightarrow \cos 2\theta ={{\cos }^{2}}\theta -\left( 1-{{\cos }^{2}}\theta \right) \\
 & \Rightarrow \cos 2\theta =2{{\cos }^{2}}\theta -1....\left( ii \right) \\
\end{align}$
Using expressions (i) and (ii) in our given expression, we will get,
 $\Rightarrow \dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }=\dfrac{\sin \theta +2\sin \theta \cos \theta }{1+\cos \theta +2{{\cos }^{2}}\theta -1}$
Taking $\sin \theta $ common from the numerator and $\cos \theta $ common from the denominator, we will get,
$\Rightarrow \dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }=\dfrac{\sin \theta \left( 1+2\cos \theta \right)}{\cos \theta \left( 1+2\cos \theta \right)}$
Cancelling out the $\left( 1+2\cos \theta \right)$ terms from the numerator and denominator, we will get,
$\Rightarrow \dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }=\dfrac{\sin \theta }{\cos \theta }$
Now, we are well aware of the fact that the ratio of $\sin \theta $ and $\cos \theta $ is another trigonometric ratio, which is called tangent, and is denoted by $\tan \theta $ . So,
$\Rightarrow \dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }=\tan \theta $
Thus, we can conclude that the given trigonometric expression $\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }$ is equal to $\tan \theta $ .

So, the correct answer is “Option C”.

Note: For solving this problem, we have started from the very basics and then concluded it. But, in the examinations, there would not be so much time available. So, we have to memorise some quick trigonometric formulae like,
$\cos 2\theta =2{{\cos }^{2}}\theta -1$ and
$\sin 2\theta =2\sin \theta \cos \theta $
Remembering these formulas will help us solve these problems in the blink of an eye.