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Mathematics

Prove the following identity:

(cos⁡A1+sin⁡A)+tan⁡A=sec⁡A\Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A = \sec A

Trigonometric Identities

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Answer

The L.H.S of above equation can be written as,

⇒(cos⁡A1+sin⁡A)+tan⁡A⇒(cos⁡A1+sin⁡A×1−sin⁡A1−sin⁡A)+sin⁡Acos⁡A⇒(cos⁡A(1−sin⁡A)1−sin⁡2A)+sin⁡Acos⁡A⇒(cos⁡A(1−sin⁡A)cos⁡2A)+sin⁡Acos⁡A⇒(1−sin⁡Acos⁡A)+sin⁡Acos⁡A⇒(1−sin⁡A+sin⁡Acos⁡A)⇒(1cos⁡A)⇒sec⁡A.\Rightarrow \Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A \\[1em] \Rightarrow \Big(\dfrac{\cos A}{1 + \sin A} \times \dfrac{1 - \sin A}{1 - \sin A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{\cos A(1 - \sin A)}{1 - \sin^2 A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{\cos A(1 - \sin A)}{\cos^2 A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{1 - \sin A}{\cos A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{1 - \sin A + \sin A}{\cos A} \Big) \\[1em] \Rightarrow \Big(\dfrac{1}{\cos A} \Big) \\[1em] \Rightarrow \sec A.

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

Hence, proved that (cos⁡A1+sin⁡A)+tan⁡A=sec⁡A\Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A = \sec A.

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