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Mathematics

Prove the following identity:

(tan⁡A1−cot⁡A)+(cot⁡A1−tan⁡A)=sec⁡Acosec⁡A+1\Big(\dfrac{\tan A}{1 - \cot A}\Big) + \Big(\dfrac{\cot A}{1 - \tan A}\Big) = \sec A \cosec A + 1

Trigonometric Identities

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Answer

Solving L.H.S of equation,

⇒sin⁡Acos⁡A1−cos⁡Asin⁡A+cos⁡Asin⁡A1−sin⁡Acos⁡A⇒sin⁡Acos⁡Asin⁡A−cos⁡Asin⁡A+cos⁡Asin⁡Acos⁡A−sin⁡Acos⁡A⇒sin⁡2Acos⁡Asin⁡A−cos⁡A+cos⁡2Asin⁡Acos⁡A−sin⁡A⇒sin⁡2Acos⁡A(sin⁡A−cos⁡A)−cos⁡2Asin⁡A(sin⁡A−cos⁡A)⇒1sin⁡A−cos⁡A(sin⁡2Acos⁡A−cos⁡2Asin⁡A)⇒1sin⁡A−cos⁡A(sin⁡3A−cos⁡3Acos⁡Asin⁡A)⇒1sin⁡A−cos⁡A((sin⁡A−cos⁡A)(sin⁡2A+cos⁡2A+sin⁡Acos⁡A)cos⁡Asin⁡A)⇒1+sin⁡Acos⁡Acos⁡Asin⁡A⇒1cos⁡Asin⁡A+cos⁡Asin⁡Acos⁡Asin⁡A⇒sec⁡Acosec⁡A+1.\Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{1 - \dfrac{\cos A}{\sin A}} + \dfrac{\dfrac{\cos A}{\sin A}}{1 - \dfrac{\sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{\sin A - \cos A}{\sin A}} + \dfrac{\dfrac{\cos A}{\sin A}}{\dfrac{\cos A - \sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\dfrac{\sin^2 A}{\cos A}}{\sin A - \cos A} + \dfrac{\dfrac{\cos^2 A}{\sin A}}{\cos A - \sin A} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A(\sin A - \cos A)} - \dfrac{\cos^2 A}{\sin A(\sin A - \cos A)} \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{\sin^2 A}{\cos A} - \dfrac{\cos^2 A}{\sin A} \Big) \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{\sin^3 A - \cos^3 A}{\cos A \sin A} \Big) \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{\cos A \sin A} \Big) \\[1em] \Rightarrow \dfrac{1 + \sin A \cos A}{\cos A \sin A} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A} + \dfrac{\cos A \sin A}{\cos A \sin A} \\[1em] \Rightarrow \sec A \cosec A + 1.

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

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

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