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Mathematics

Prove the following identity:

(cos⁡A1−tan⁡A)+(sin⁡2Asin⁡A−cos⁡A)=cos⁡A+sin⁡A\Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) = \cos A + \sin A

Trigonometric Identities

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Answer

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

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

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

Hence, proved that (cos⁡A1−tan⁡A)+(sin⁡2Asin⁡A−cos⁡A)=cos⁡A+sin⁡A\Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) = \cos A + \sin A.

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