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Question: Prove the following \(1 + \dfrac{{\tan A}}{{\sin A}} + 1 + \dfrac{{\cot A}}{{\cos A}} = 2\cos ecA\) ...

Prove the following 1+tan⁡Asin⁡A+1+cot⁡Acos⁡A=2cos⁡ecA1 + \dfrac{{\tan A}}{{\sin A}} + 1 + \dfrac{{\cot A}}{{\cos A}} = 2\cos ecA ?

Explanation

Solution

Here we are given a trigonometric equation and we need to verify whether the given equation is true or not. To prove the equation, we need to solve the left-hand side equation. Then we have to check whether the obtained result from the left-hand side of the equation is as same as the given right-hand side of the equation. That is, if the answer from the left-hand side is 2cos⁡ecA2\cos ecA, then the given equation is proved, or else the given equation is a false statement.
Formula to be used:
a) The following reciprocal identities are to be used.
1sin⁡θ=cos⁡ecθ\dfrac{1}{{\sin \theta }} = \cos ec\theta
1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta
b) The following quotient identities are to be used.
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}

Complete answer:
The given trigonometric equation is 1+tan⁡Asin⁡A+1+cot⁡Acos⁡A=2cos⁡ecA1 + \dfrac{{\tan A}}{{\sin A}} + 1 + \dfrac{{\cot A}}{{\cos A}} = 2\cos ecA
First, let us consider and solve the left-hand side of the equation.
1+tan⁡Asin⁡A+1+cot⁡Acos⁡A\dfrac{{1 + \tan A}}{{\sin A}} + \dfrac{{1 + \cot A}}{{\cos A}}
Now we shall split all the terms present on the numerators.
1+tan⁡Asin⁡A+1+cot⁡Acos⁡A=1sin⁡A+tan⁡Asin⁡A+1cos⁡A+cot⁡Acos⁡A\dfrac{{1 + \tan A}}{{\sin A}} + \dfrac{{1 + \cot A}}{{\cos A}} = \dfrac{1}{{\sin A}} + \dfrac{{\tan A}}{{\sin A}} + \dfrac{1}{{\cos A}} + \dfrac{{\cot A}}{{\cos A}}
=cos⁡ecA+tan⁡Asin⁡A+sec⁡A+cot⁡Acos⁡A= \cos ecA + \dfrac{{\tan A}}{{\sin A}} + \sec A + \dfrac{{\cot A}}{{\cos A}} (Here we have applied the reciprocal identities) =cos⁡ecA+tan⁡A1sin⁡A+sec⁡A+cot⁡A1cos⁡A = \cos ecA + \tan A\dfrac{1}{{\sin A}} + \sec A + \cot A\dfrac{1}{{\cos A}}
=cos⁡ecA+sin⁡Acos⁡A1sin⁡A+sec⁡A+cos⁡Asin⁡A1cos⁡A= \cos ecA + \dfrac{{\sin A}}{{\cos A}}\dfrac{1}{{\sin A}} + \sec A + \dfrac{{\cos A}}{{\sin A}}\dfrac{1}{{\cos A}} (Now we have applied the quotient identities)
=cos⁡ecA+1cos⁡A+sec⁡A+1sin⁡A= \cos ecA + \dfrac{1}{{\cos A}} + \sec A + \dfrac{1}{{\sin A}}
(Here we have applied the reciprocal identities; 1sin⁡θ=cos⁡ecθ\dfrac{1}{{\sin \theta }} = \cos ec\theta and 1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta )
=cos⁡ecA+sec⁡A+sec⁡A+cos⁡ecA= \cos ecA + \sec A + \sec A + \cos ecA
=2cos⁡ecA+2sec⁡A= 2\cos ecA + 2\sec A
=2(cos⁡ecA+sec⁡A)= 2\left( {\cos ecA + \sec A} \right)
=2(secA+cosec⁡A)= 2\left( {secA + co\sec A} \right)
We know that if the answer from the left-hand side is 2cos⁡ecA2\cos ecA, then the given equation is proved, or else the given equation is a false statement.
Since the answer mismatches the right-hand side, the equation is a false statement.

Note:
We can also solve the right-hand side of the given equation but it’s quite hard. We will get the false result as we get earlier. If the given equation is trigonometric, then we need to analyze the equation what are the trigonometric identities it will need and so we can able to solve the problem.