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Question: Simplify the given expression and get the value of ? in \[\left[ \dfrac{\sin 2A}{1+\cos 2A} \right]\...

Simplify the given expression and get the value of ? in [sin⁡2A1+cos⁡2A][cos⁡A1+cos⁡A]=?\left[ \dfrac{\sin 2A}{1+\cos 2A} \right]\left[ \dfrac{\cos A}{1+\cos A} \right]=?
A. tan⁡A2\tan \dfrac{A}{2}
B. cot⁡A2\cot \dfrac{A}{2}
C. sec⁡A2\sec \dfrac{A}{2}
D. cosecA2\text{cosec}\dfrac{A}{2}

Explanation

Solution

In this problem, we have to simplify and find the value of the given trigonometric expression. Here we have to use suitable trigonometric identities and simplify them step by step. We can first use the identity cos⁡2A=2cos⁡2A−1\cos 2A=2{{\cos }^{2}}A-1, we can then cancel the similar terms and we can use another identities sin⁡A=2tan⁡A21+tan⁡2A2,cos⁡A=1−tan⁡2A21+tan⁡2A2\sin A=\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}},\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}} and simplify it to get the required answer.

Complete step by step solution:
Here we have to find the value for the given trigonometric expression
[sin⁡2A1+cos⁡2A][cos⁡A1+cos⁡A]\left[ \dfrac{\sin 2A}{1+\cos 2A} \right]\left[ \dfrac{\cos A}{1+\cos A} \right]
We can first substitute 1+cos⁡2A=2cos⁡2A1+\cos 2A=2{{\cos }^{2}}A in the above step, we get
=[sin⁡2A2cos⁡2A][cos⁡A1+cos⁡A]= \left[ \dfrac{\sin 2A}{2{{\cos }^{2}}A} \right]\left[ \dfrac{\cos A}{1+\cos A} \right]
We can now cancel similar terms in the above step, we get
=[sin⁡2A2cos⁡A][11+cos⁡A]= \left[ \dfrac{\sin 2A}{2\cos A} \right]\left[ \dfrac{1}{1+\cos A} \right]
We can now substitute the identity sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A in the above step, we get
=[2sin⁡Acos⁡A2cos⁡A][11+cos⁡A]= \left[ \dfrac{2\sin A\cos A}{2\cos A} \right]\left[ \dfrac{1}{1+\cos A} \right]
We can now cancel the similar terms, we get
=sin⁡A1+cos⁡A= \dfrac{\sin A}{1+\cos A}
We can now substitute the identities sin⁡A=2tan⁡A21+tan⁡2A2,cos⁡A=1−tan⁡2A21+tan⁡2A2\sin A=\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}},\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}} in the above step⇒2tan⁡A21+tan⁡2A21+1−tan⁡2A21+tan⁡2A2\Rightarrow \dfrac{\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}}{1+\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}}
We can now cross multiply the terms in the denominator, we get
=2tan⁡A21+tan⁡2A21+tan⁡2A2+1−tan⁡2A21+tan⁡2A2=2tan⁡A21+tan⁡2A221+tan⁡2A2= \dfrac{\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}}{\dfrac{1+{{\tan }^{2}}\dfrac{A}{2}+1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}}=\dfrac{\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}}{\dfrac{2}{1+{{\tan }^{2}}\dfrac{A}{2}}}
We can now take reciprocal in the above step, we get
=2tan⁡A21+tan⁡2A2×1+tan⁡2A22= \dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}\times \dfrac{1+{{\tan }^{2}}\dfrac{A}{2}}{2}
We can now cancel the similar terms in the above step, we get
=tan⁡A2= \tan \dfrac{A}{2}
Therefore, the answer is option A. tan⁡A2\tan \dfrac{A}{2}

Note: We should always remember some of the trigonometric identities to be substituted in the given expression in order to simplify the substituted terms step by step and to find the required answer. We should remember some of the identities such as sin⁡A=2tan⁡A21+tan⁡2A2,cos⁡A=1−tan⁡2A21+tan⁡2A2\sin A=\dfrac{2\tan \dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}},\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}} and sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A. We should concentrate while simplifying the fraction terms, we can use the reciprocal method to solve in an easy way.