Prove that : \(\frac{1}{cosec θ- cot θ}\) = (cosecθ + cotθ).

1.	Prove that : \(\frac{1}{cosec θ- cot θ}\) = (cosecθ + cotθ).
Answer» L.H.S = \(\frac{1}{cosec θ+ cot θ}\) = \(\frac{cosec\theta+cot\theta}{(cosecθ − cotθ)(cosecθ + cotθ)}\) (By multiplying numerator and denominator by cosecθ + cotθ) = \(\frac{cosecθ + cotθ}{cosec^2θ − cot^2θ}\) (∵ (a+b)(a–b) = a² − b² ) = \(\frac{cosecθ + cotθ }{1}\) = cosecθ + cotθ = R.H.S. (∵ cosec²θ = 1 + cot²θ ⇒ cosec²θ − cot²θ = 1) Hence proved

Answer»

L.H.S = \(\frac{1}{cosec θ+ cot θ}\) = \(\frac{cosec\theta+cot\theta}{(cosecθ − cotθ)(cosecθ + cotθ)}\)

(By multiplying numerator and denominator by cosecθ + cotθ)

= \(\frac{cosecθ + cotθ}{cosec^2θ − cot^2θ}\) (∵ (a+b)(a–b) = a² − b² )

= \(\frac{cosecθ + cotθ }{1}\) = cosecθ + cotθ = R.H.S.

(∵ cosec²θ = 1 + cot²θ ⇒ cosec²θ − cot²θ = 1)

Hence proved

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Prove that : \(\frac{1}{cosec θ- cot θ}\) = (cosecθ + cotθ).

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