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Mathematics

Prove the following identity:

(cot⁡θ+cosec⁡θ−1cot⁡θ−cosec⁡θ+1)=(1+cos⁡θsin⁡θ)\Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big)

Trigonometric Identities

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Answer

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

⇒cot⁡θ+cosec⁡θ−1cot⁡θ−cosec⁡θ+1⇒cos⁡θsin⁡θ+1sin⁡θ−1cos⁡θsin⁡θ−1sin⁡θ+1⇒cos⁡θ+1−sin⁡θsin⁡θcos⁡θ−1+sin⁡θsin⁡θ⇒cos⁡θ+1−sin⁡θcos⁡θ−1+sin⁡θ⇒cos⁡θ+(1−sin⁡θ)cos⁡θ−(1−sin⁡θ)⇒cos⁡θ+(1−sin⁡θ)cos⁡θ−(1−sin⁡θ)×cos⁡θ+(1−sin⁡θ)cos⁡θ+(1−sin⁡θ)⇒[cos⁡θ+(1−sin⁡θ)]2cos⁡2θ−(1−sin⁡θ)2⇒cos⁡2θ+(1−sin⁡θ)2+2cos⁡θ(1−sin⁡θ)cos⁡2θ−(1−sin⁡θ)2⇒cos⁡2θ+sin⁡2θ+1+2cos⁡θ−2sin⁡θ−2sin⁡θcos⁡θcos⁡2θ−1−sin⁡2θ+2sin⁡θ By formula, sin⁡2A+cos⁡2A=1⇒1+1+2cos⁡θ−2sin⁡θ−2sin⁡θcos⁡θ1−sin⁡2θ−1−sin⁡2θ+2sin⁡θ⇒2+2cos⁡θ−2sin⁡θ−2sin⁡θcos⁡θ2sin⁡θ−2sin⁡2θ⇒2(1+cos⁡θ)−2sin⁡θ(1+cos⁡θ)2sin⁡θ(1−sin⁡θ)⇒(1+cos⁡θ)(2−2sin⁡θ)2sin⁡θ(1−sin⁡θ)⇒2(1+cos⁡θ)(1−sin⁡θ)2sin⁡θ(1−sin⁡θ)⇒1+cos⁡θsin⁡θ.\Rightarrow \dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\cos \theta}{\sin \theta} + \dfrac{1}{\sin \theta} - 1}{\dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\cos \theta + 1 - \sin \theta}{\sin \theta}}{\dfrac{\cos \theta - 1 + \sin \theta}{\sin \theta}} \\[1em] \Rightarrow \dfrac{\cos \theta + 1 - \sin \theta}{\cos \theta - 1 + \sin \theta} \\[1em] \Rightarrow \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta - (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta - (1 - \sin \theta)} \times \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta + (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{[\cos \theta + (1 - \sin \theta)]^2}{\cos^2 \theta - (1 - \sin \theta)^2} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + (1 - \sin \theta)^2 + 2\cos \theta(1 - \sin \theta)}{\cos^2 \theta - (1 - \sin \theta)^2} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + \sin^2 \theta + 1 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{\cos^2 \theta - 1 - \sin^2 \theta + 2\sin \theta} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1 + 1 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{1 - \sin^2 \theta - 1 - \sin^2 \theta + 2\sin \theta} \\[1em] \Rightarrow \dfrac{2 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{2\sin \theta - 2\sin^2 \theta} \\[1em] \Rightarrow \dfrac{2(1 + \cos \theta) - 2\sin \theta (1 + \cos \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{(1 + \cos \theta) (2 - 2\sin \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{2(1 + \cos \theta) (1 - \sin \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{1 + \cos \theta}{\sin \theta}.

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

Hence, proved that (cot⁡θ+cosec⁡θ−1cot⁡θ−cosec⁡θ+1)=(1+cos⁡θsin⁡θ)\Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big).

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