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Mathematics

Prove the following identity:

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

Trigonometric Identities

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Answer

The equation can be written as,

1sec⁡A+tan⁡A+1sec⁡A−tan⁡A=2cos⁡A\dfrac{1}{\sec A + \tan A} + \dfrac{1}{\sec A - \tan A} = \dfrac{2}{\cos A}

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

⇒sec⁡A−tan⁡A+sec⁡A+tan⁡A(sec⁡A+tan⁡A)(sec⁡A−tan⁡A)⇒2sec⁡A(sec⁡2A−tan⁡2A)⇒2sec⁡A⇒2cos⁡A.\Rightarrow \dfrac{\sec A - \tan A + \sec A + \tan A}{(\sec A + \tan A)(\sec A - \tan A)} \\[1em] \Rightarrow \dfrac{2\sec A}{(\sec^2 A - \tan^2 A)} \\[1em] \Rightarrow 2\sec A \\[1em] \Rightarrow \dfrac{2}{\cos A}.

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

Hence, proved that

(1sec⁡A+tan⁡A)−(1cos⁡A)=(1cos⁡A)−(1sec⁡A−tan⁡A)\Big(\dfrac{1}{\sec A + \tan A}\Big) - \Big(\dfrac{1}{\cos A}\Big) = \Big(\dfrac{1}{\cos A}\Big) - \Big(\dfrac{1}{\sec A - \tan A}\Big).

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