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Mathematics

Prove the following identity:

(1 + cot A - cosec A)(1 + tan A + sec A) = 2

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

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

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

Hence, proved that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.

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