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Mathematics

Prove the following identity:

(cosec⁡Acosec⁡A−1)+(cosec⁡Acosec⁡A+1)=2sec⁡2A\Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒cosec⁡Acosec⁡A−1+cosec⁡Acosec⁡A+1⇒cosec⁡A(cosec⁡A+1)+cosec⁡A(cosec⁡A−1)(cosec⁡A−1)(cosec⁡A+1)⇒cosec⁡2A+cosec⁡A+cosec⁡2A−cosec⁡Acosec⁡2A−1⇒2cosec⁡2Acot⁡2A⇒2×1sin⁡2Acos⁡2Asin⁡2A⇒2cos⁡2A⇒2sec⁡2A.\Rightarrow \dfrac{\cosec A}{\cosec A - 1} + \dfrac{\cosec A}{\cosec A + 1} \\[1em] \Rightarrow \dfrac{\cosec A(\cosec A + 1) + \cosec A(\cosec A - 1)}{(\cosec A - 1) (\cosec A + 1)} \\[1em] \Rightarrow \dfrac{\cosec^2 A + \cosec A + \cosec^2 A - \cosec A}{\cosec^2 A - 1} \\[1em] \Rightarrow \dfrac{2\cosec^2 A}{\cot^2 A} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\sin^2 A}}{\dfrac{\cos^2 A}{\sin^2 A}} \\[1em] \Rightarrow \dfrac{2}{\cos^2 A} \\[1em] \Rightarrow 2\sec^2 A.

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

Hence, proved that (cosec⁡Acosec⁡A−1)+(cosec⁡Acosec⁡A+1)=2sec⁡2A\Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A.

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