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Mathematics

Prove the following identity:

(tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1

Trigonometric Identities

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Answer

Solving L.H.S. of the above equation :

⇒ (tan A + cot A)(cosec A - sin A)(sec A - cos A)

⇒(sin⁡Acos⁡A+cos⁡Asin⁡A)(1sin⁡A−sin⁡A)(1cos⁡A−cos⁡A)⇒(sin⁡2A+cos⁡2Acos⁡Asin⁡A)(1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)By formula,sin⁡2A+cos⁡2A=1,1−sin⁡2A=cos⁡2Aand1−cos⁡2A=sin⁡2A.⇒(1cos⁡Asin⁡A)(cos⁡2Asin⁡A)(sin⁡2Acos⁡A)⇒(cos⁡2Asin⁡2Acos⁡2Asin⁡2A)⇒1.\Rightarrow \Big(\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \Big) \Big(\dfrac{1}{\sin A} - \sin A\Big) \Big(\dfrac{1}{\cos A} - \cos A \Big) \\[1em] \Rightarrow \Big(\dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \Big) \Big(\dfrac{1 - \sin^2 A}{\sin A}\Big) \Big(\dfrac{1 - \cos^2 A}{\cos A}\Big) \\[1em] \text{By formula,} \sin^2 A + \cos^2 A = 1, 1 - \sin^2 A = \cos^2 A and 1 - \cos^2 A = \sin^2 A. \\[1em] \Rightarrow \Big(\dfrac{1}{\cos A \sin A} \Big) \Big(\dfrac{\cos^2 A}{\sin A}\Big) \Big(\dfrac{\sin^2 A}{\cos A}\Big) \\[1em] \Rightarrow \Big(\dfrac{\cos^2 A\sin^2 A}{\cos^2 A \sin^2 A} \Big) \\[1em] \Rightarrow 1.

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

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

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