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Mathematics

Prove the following identity:

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

Trigonometric Identities

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Answer

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

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

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

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

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