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Mathematics

Prove the following identity:

(cot⁡A−12−sec⁡2A)=(cot⁡A1+tan⁡A)\Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big)

Trigonometric Identities

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Answer

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

⇒cos⁡Asin⁡A−12−1cos⁡2A⇒cos⁡A−sin⁡Asin⁡A2cos⁡2−1cos⁡2A⇒cos⁡2A(cos⁡A−sin⁡A)sin⁡A(2cos⁡2A−1)⇒cos⁡2A(cos⁡A−sin⁡A)sin⁡A[2cos⁡2A−(sin⁡2A+cos⁡2A)]⇒cos⁡2A(cos⁡A−sin⁡A)sin⁡A[2cos⁡2A−sin⁡2A−cos⁡2A]⇒cos⁡2A(cos⁡A−sin⁡A)sin⁡A(cos⁡2A−sin⁡2A)⇒cos⁡2A(cos⁡A−sin⁡A)sin⁡A(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)⇒cos⁡2Asin⁡A(cos⁡A+sin⁡A)⇒(cos⁡A)(cos⁡A)sin⁡A(cos⁡A+sin⁡A)⇒cot⁡A(cos⁡A)(cos⁡A+sin⁡A)⇒cot⁡A(cos⁡A)cos⁡A(cos⁡A+sin⁡A)cos⁡A⇒cot⁡A1+tan⁡A.\Rightarrow \dfrac{\dfrac{\cos A}{\sin A} - 1}{2 - \dfrac{1}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{\dfrac{\cos A - \sin A}{\sin A}}{\dfrac{2\cos^2 - 1}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(2\cos^2 A - 1)} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A[2\cos^2 A - (\sin^2 A + \cos^2 A)]} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A[2\cos^2 A - \sin^2 A - \cos^2 A]} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(\cos^2 A - \sin^2 A)} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(\cos A - \sin A)(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\cos^2 A}{ \sin A(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{(\cos A)(\cos A)}{ \sin A(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\cot A(\cos A)}{(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\dfrac{\cot A(\cos A)}{\cos A}}{\dfrac{(\cos A + \sin A)}{\cos A}} \\[1em] \Rightarrow \dfrac{\cot A}{1 + \tan A}.

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

Hence, proved that (cot⁡A−12−sec⁡2A)=(cot⁡A1+tan⁡A)\Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big).

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