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Mathematics

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

(cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}

Trigonometric Identities

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Answer

Given,

Equation : (cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}

Solving L.H.S. of the above equation :

⇒(1sin θ−cos θsin θ)2⇒(1−cos θsin θ)2⇒(1 - cos θ)2sin2θ⇒(1 - cos θ)21−cos2θ⇒(1 - cos θ)2(1 - cos θ)(1 + cos θ)⇒1 - cos θ1 + cos θ.\Rightarrow \Big(\dfrac{1}{\text{sin θ}} - \dfrac{\text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{sin}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{1 - \text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{(1 - cos θ)(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}.

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

Hence, proved that (cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}.

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