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Question: Simplify the following trigonometric expression. \(\dfrac{\sin 2\theta }{1+\cos 2\theta }\)....

Simplify the following trigonometric expression.
sin⁡2θ1+cos⁡2θ\dfrac{\sin 2\theta }{1+\cos 2\theta }.

Explanation

Solution

Hint: To solve this problem, we should be aware about the basic properties of trigonometric terms involving multiple angles. Thus, we should know that-
cos⁡2θ=2cos⁡2θ−1 sin⁡2θ=2sin⁡θcos⁡θ \begin{aligned} & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ & \sin 2\theta =2\sin \theta \cos \theta \\\ \end{aligned}
These expressions would be helpful in solving the above mentioned problem.

Complete step-by-step answer:
Before we begin solving the question, it is important to know about multiple angles in relation to trigonometry. Generally, to find the expression of sin⁡2θ\sin 2\theta , we express 2θ=θ+θ2\theta =\theta +\theta and then use the property that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A+B)=\sin A\cos B+\cos A\sin B. Thus, in this case, A=B=θ\theta . Thus, we have,
sin⁡(θ+θ)=sin⁡θcos⁡θ+sin⁡θcos⁡θ\sin (\theta +\theta )=\sin \theta \cos \theta +\sin \theta \cos \theta
sin⁡(2θ)=2sin⁡θcos⁡θ\sin (2\theta )=2\sin \theta \cos \theta
We can similarly derive the formula for cos⁡2θ\cos 2\theta by a similar procedure. We use the property that cos(A+B) = cosAcosB – sinAsinB. Thus, in this case, A=B=θ\theta . Thus, we have,
cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ cos⁡(2θ)=cos⁡2θ−sin⁡2θ cos⁡(2θ)=cos⁡2θ−(1−cos⁡2θ) cos⁡(2θ)=2cos⁡2θ−1 \begin{aligned} & \cos (\theta +\theta )=\cos \theta \cos \theta -\sin \theta \sin \theta \\\ & \cos (2\theta )={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ & \cos (2\theta )={{\cos }^{2}}\theta -(1-{{\cos }^{2}}\theta ) \\\ & \cos (2\theta )=2{{\cos }^{2}}\theta -1 \\\ \end{aligned}

Thus, we make use of these properties to solve the above problem. Now, we have,
sin⁡2θ1+cos⁡2θ\dfrac{\sin 2\theta }{1+\cos 2\theta }

& =\dfrac{2\sin \theta \cos \theta }{1+2{{\cos }^{2}}\theta -1} \\\ & =\dfrac{2\sin \theta \cos \theta }{2{{\cos }^{2}}\theta } \\\ & =\dfrac{\sin \theta }{\cos \theta } \\\ & =\tan \theta \\\ \end{aligned}$$ Hence, the correct answer after simplification is tan$\theta $. Note: An alternative way to solve this problem is to express denominator and numerator in terms of tan$\theta $ and then solve. Thus, we have, =$\dfrac{\sin 2\theta }{1+\cos 2\theta }$ Now, to get expression in terms of tan$\theta $, we divide the numerator and denominator by cos$2\theta $, thus we get, =$\dfrac{\tan 2\theta }{1+\sec 2\theta }$ Since, $\dfrac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta $ and $\dfrac{1}{\cos 2\theta }=\sec 2\theta $. By using the property that $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and $\sec 2\theta =\dfrac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }$, we get, $\begin{aligned} & =\dfrac{\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }}{1+\dfrac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }} \\\ & \\\ & =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta +1+{{\tan }^{2}}\theta } \\\ & \\\ & =\dfrac{2\tan \theta }{2} \\\ & \\\ & =\tan \theta \\\ \end{aligned}$ Thus, we were able to arrive at the same solution using a different technique. Thus, in question involving solving a trigonometric expression, there are generally multiple ways to arrive at the solution, however, at times it is generally beneficial to use one method over another due to its ease.