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Mathematics

Prove the following identity:

(1+sin⁡Acos⁡A)+(cos⁡A1+sin⁡A)=2sec⁡A\Big(\dfrac{1 + \sin A}{\cos A}\Big) + \Big(\dfrac{\cos A}{1 + \sin A}\Big) = 2 \sec A

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒1+sin⁡Acos⁡A+cos⁡A1+sin⁡A⇒(1+sin⁡A)2+cos⁡2Acos⁡A(1+sin⁡A)⇒1+2sin⁡A+sin⁡2A+cos⁡2Acos⁡A(1+sin⁡A) By formula, sin⁡2A+cos⁡2A=1⇒1+2sin⁡A+1cos⁡A(1+sin⁡A)⇒2+2sin⁡Acos⁡A(1+sin⁡A)⇒2(1+sin⁡A)cos⁡A(1+sin⁡A)⇒2cos⁡A⇒2sec⁡A.\Rightarrow \dfrac{1 + \sin A}{\cos A} + \dfrac{\cos A}{1 + \sin A} \\[1em] \Rightarrow \dfrac{(1 + \sin A)^2 + \cos^2 A}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{1 + 2\sin A + \sin^2 A + \cos^2 A}{\cos A(1 + \sin A)} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1 + 2\sin A + 1}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2 + 2\sin A}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2(1 + \sin A)}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2}{\cos A} \\[1em] \Rightarrow 2\sec A.

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

Hence, proved that (1+sin⁡Acos⁡A)+(cos⁡A1+sin⁡A)=2sec⁡A\Big(\dfrac{1 + \sin A}{\cos A}\Big) + \Big(\dfrac{\cos A}{1 + \sin A}\Big) = 2 \sec A.

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