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Question: Let A is an angle of a triangle then \(\cos ecA - 2\cot 2A\cos A\) \(\cos ecA - 2\cot 2A\cos A\) is ...

Let A is an angle of a triangle then cos⁡ecA−2cot⁡2Acos⁡A\cos ecA - 2\cot 2A\cos A cos⁡ecA−2cot⁡2Acos⁡A\cos ecA - 2\cot 2A\cos A is equal to which option?
A.) 2sin⁡A2\sin A
B.) sec⁡A\sec A
C.) 2cos⁡Acot⁡A2\cos A\cot A
D.) None of this

Explanation

Solution

In order to solve the above problem, You need to know about some basic and fundamental formulas of trigonometric equations. As you see, this question will need some additional formula to get the final answer. You have to use the below formula to get the final answer.
(1−cos⁡A) 2=sin⁡2A\dfrac{{(1 - \cos A)\,}}{2} = {\sin ^2}A for converting cos⁡2A\cos 2Ainto sin⁡2A{\sin ^2}Aterms for solve and use sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A and also cot⁡2A=cos⁡2Asin⁡2A\cot 2A = \dfrac{{\cos 2A}}{{\sin 2A}}

Complete answer:
First of all we have to write our given equation,
⇒cos⁡ecA−2cot⁡2Acos⁡A\Rightarrow \cos ecA - 2\cot 2A\cos A
Now, Convert cot⁡2A\cot 2A to cos⁡2Asin⁡2A\dfrac{{\cos 2A}}{{\sin 2A}} and put these values in our given question which is cos⁡ecA−2cot⁡2Acos⁡A\cos ecA - 2\cot 2A\cos Aand take inverse of cosec so it becomes sine function,
After replacement you will get this kind of equation.
⇒1sin⁡A−2(cos⁡2Asin⁡2A)cos⁡A\Rightarrow \dfrac{1}{{\sin A}} - 2(\dfrac{{\cos 2A}}{{\sin 2A}})\cos A
Now, let’s convert above term and we will get,
⇒sin⁡2A=2sin⁡Acos⁡A\Rightarrow \sin 2A = 2\sin A\cos A
Put values of sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos Ainto the above equation so that we can move further,
⇒1sin⁡A−2(cos⁡2A2sin⁡Acos⁡A)cos⁡A\Rightarrow \dfrac{1}{{\sin A}} - 2(\dfrac{{\cos 2A}}{{2\sin A\cos A}})\cos A
Now, do cancelation in numerator and denominator and we will get following term,
⇒1sin⁡A−(cos⁡2Asin⁡A)\Rightarrow \dfrac{1}{{\sin A}} - (\dfrac{{\cos 2A}}{{\sin A}})
Do some more simplification and we will get,
⇒1−cos⁡2Asin⁡A\Rightarrow \dfrac{{1 - \cos 2A}}{{\sin A}}
Now, let’s convert above equation (1−cos⁡A) 2=sin⁡2A\dfrac{{(1 - \cos A)\,}}{2} = {\sin ^2}A
⇒2sin⁡2Asin⁡A\Rightarrow \dfrac{{2{{\sin }^2}A}}{{\sin A}} ___equation (3)
Do cancelation in numerator and denominator and we will get following term,
⇒2sin⁡A\Rightarrow 2\sin A
Hence, the correct option is (A) .

Note:
This problem can be also solved by the hit and trial method. First you will be putting some random value of trigonometric angle. After that you also need to put that same value in each option. And after that compare every value with the given equation value so you will find the answer very quickly.