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Mathematics

Prove the following identity:

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

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

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

⇒(1sin⁡θ−sin⁡θ)(1cos⁡θ−cos⁡θ)(sin⁡θcos⁡θ+cos⁡θsin⁡θ)⇒(1−sin⁡2θsin⁡θ)(1−cos⁡2θcos⁡θ)(sin⁡2θ+cos⁡2θcos⁡θsin⁡θ)⇒(cos⁡2θsin⁡θ)(sin⁡2θcos⁡θ)(1cos⁡θsin⁡θ)⇒cos⁡2θsin⁡2θcos⁡2θsin⁡2θ⇒1.\Rightarrow \Big(\dfrac{1}{\sin \theta } - \sin \theta \Big) \Big(\dfrac{1}{\cos \theta } - \cos \theta \Big) \Big(\dfrac{\sin \theta}{\cos \theta } + \dfrac{\cos \theta}{\sin \theta }\Big) \\[1em] \Rightarrow \Big(\dfrac{1 - \sin^2 \theta}{\sin \theta}\Big) \Big(\dfrac{1 - \cos^2 \theta}{\cos \theta } \Big) \Big(\dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta } \Big) \\[1em] \Rightarrow \Big(\dfrac{\cos^2 \theta}{\sin \theta}\Big) \Big(\dfrac{\sin^2 \theta}{\cos \theta } \Big) \Big(\dfrac{1}{\cos \theta \sin \theta } \Big) \\[1em] \Rightarrow \dfrac{\cos^2 \theta \sin^2 \theta}{\cos^2 \theta \sin^2 \theta } \\[1em] \Rightarrow 1.

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

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

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