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Mathematics

Prove the following identities :

1 - sin A1 + sin A\dfrac{\text{1 - sin A}}{\text{1 + sin A}} = (sec A - tan A)2

Trigonometric Identities

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Answer

Solving R.H.S. of the equation :

⇒(sec A - tan A)2⇒(1cos A−sin Acos A)2⇒(1 - sin Acos A)2⇒(1 - sin A)2cos2A\Rightarrow \text{(sec A - tan A)}^2 \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{cos}^2 A}

By formula,

cos2 A = 1 - sin2 A

⇒(1 - sin A)21 - sin2A⇒(1 - sin A)2(1 - sin A)(1 + sin A)⇒1 - sin A1 + sin A.\Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{1 - sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A}}{\text{1 + sin A}}.

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

Hence, proved that 1 - sin A1 + sin A\dfrac{\text{1 - sin A}}{\text{1 + sin A}} = (sec A - tan A)2.

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