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Mathematics

Prove the following identities :

1cos A + sin A+1cos A - sin A=2 cos A2 cos2A−1\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒1cos A + sin A+1cos A - sin A⇒cos A - sin A + cos A + sin Acos2A−sin2A⇒2 cos Acos2A−sin2A\Rightarrow \dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} \\[1em] \Rightarrow \dfrac{\text{cos A - sin A + cos A + sin A}}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - \text{sin}^2 A}

By formula,

sin2 A = 1 - cos2 A

⇒2 cos Acos2A−(1−cos2A)⇒2 cos A2 cos2A−1.\Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - (1 - \text{cos}^2 A)} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}.

Since, L.H.S. = R.H.S.

Hence, proved that 1cos A + sin A+1cos A - sin A=2 cos A2 cos2A−1\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}.

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