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Mathematics

Prove the following identities :

1 - cos2A1 + sin A\dfrac{\text{cos}^2 A}{\text{1 + sin A}} = sin A

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒1−cos2A1 + sin A⇒1 + sin A - cos2A1 + sin A\Rightarrow 1 - \dfrac{\text{cos}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A - cos}^2 A}{\text{1 + sin A}}

By formula,

cos2 A = 1 - sin2 A

⇒1 + sin A - (1 - sin2A)1 + sin A⇒1 + sin A - 1 + sin2A1 + sin A⇒sin A + sin2A1 + sin A⇒sin A(1 + sin A)1 + sin A⇒sin A.\Rightarrow \dfrac{\text{1 + sin A - (1 - sin}^2 A)}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A - 1 + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A(1 + sin A)}}{\text{1 + sin A}} \\[1em] \Rightarrow \text{sin A}.

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

Hence, proved that 1 - cos2A1 + sin A\dfrac{\text{cos}^2 A}{\text{1 + sin A}} = sin A.

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