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Mathematics

Prove the following identities :

cos θ cot θ1 + sin θ\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}} = cosec θ - 1

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒cos θ×cos θsin θ(1 + sin θ)⇒cos2θsin θ(1 + sin θ)\Rightarrow \dfrac{\text{cos θ} \times \dfrac{\text{cos θ}}{\text{sin θ}}}{\text{(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{sin θ(1 + sin θ)}}

By formula,

cos2 θ = 1 - sin2 θ

⇒1 - sin2θsin θ(1 + sin θ)⇒(1 - sin θ)(1 + sin θ)sin θ(1 + sin θ)⇒1 - sin θsin θ⇒1sin θ−sin θsin θ⇒cosec θ - 1\Rightarrow \dfrac{\text{1 - sin}^2 θ}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin θ)(1 + sin θ)}}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \dfrac{1}{\text{sin θ}} - \dfrac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \text{cosec θ - 1}

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

Hence, proved that cos θ cot θ1 + sin θ\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}} = cosec θ - 1.

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