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Question: Find the value of \(\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }\) A. \(\dfrac...

Find the value of sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }
A. 12tan⁡θ\dfrac{1}{2}\tan \theta
B. 12cot⁡θ\dfrac{1}{2}\cot \theta
C. tan⁡θ\tan \theta
D. cot⁡θ\cot \theta

Explanation

Solution

At first, we find the value of sin⁡2θ\sin 2\theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . We then find the value of cos⁡2θ\cos 2\theta in terms of cos⁡θ\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,
sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }
We know the trigonometric identities which say that,
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B and
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\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 θ\theta , then, we will get,
sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ ⇒sin⁡2θ=2sin⁡θcos⁡θ....(i) \begin{aligned} & \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{aligned}
Doing the same for cos⁡(A+B)\cos \left( A+B \right) , we will get,
cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ ⇒cos⁡2θ=cos⁡2θ−sin⁡2θ \begin{aligned} & \cos \left( \theta +\theta \right)=\cos \theta \cos \theta -\sin \theta \sin \theta \\\ & \Rightarrow \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ \end{aligned}
We also know the trigonometric identity which states that,
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Using the above trigonometric identity in the equation for cos⁡2θ\cos 2\theta , we will get,
⇒cos⁡2θ=cos⁡2θ−(1−cos⁡2θ) ⇒cos⁡2θ=2cos⁡2θ−1....(ii) \begin{aligned} & \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{aligned}
Using expressions (i) and (ii) in our given expression, we will get,
⇒sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=sin⁡θ+2sin⁡θcos⁡θ1+cos⁡θ+2cos⁡2θ−1\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⁡θ\sin \theta common from the numerator and cos⁡θ\cos \theta common from the denominator, we will get,
⇒sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=sin⁡θ(1+2cos⁡θ)cos⁡θ(1+2cos⁡θ)\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 (1+2cos⁡θ)\left( 1+2\cos \theta \right) terms from the numerator and denominator, we will get,
⇒sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=sin⁡θcos⁡θ\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⁡θ\sin \theta and cos⁡θ\cos \theta is another trigonometric ratio, which is called tangent, and is denoted by tan⁡θ\tan \theta . So,
⇒sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ\Rightarrow \dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta }=\tan \theta
Thus, we can conclude that the given trigonometric expression sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ\dfrac{\sin \theta +\sin 2\theta }{1+\cos \theta +\cos 2\theta } is equal to tan⁡θ\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⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1 and
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
Remembering these formulas will help us solve these problems in the blink of an eye.