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Mathematics

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

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

Trigonometric Identities

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Answer

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

⇒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)]2cos2θ−(1 - sin θ)2⇒cos2 θ+(1 - sin θ)2+2cos θ(1 - sin θ)cos2 θ−(1+sin2θ−2sin θ)⇒cos2θ+sin2θ+1+2cos θ - 2sin θ - 2 sin θ cos θcos2 θ−1−sin2θ+2sin θ⇒1+1+2cos θ - 2sin θ - 2 sin θ cos θ1−sin2θ−1−sin2θ+2 sin θ⇒2+2cos θ - 2sin θ - 2sin θ cos θ2sin θ - 2sin2θ⇒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{\dfrac{\text{cos θ}}{\text{sin θ}} + \dfrac{1}{\text{sin θ}} - 1}{\dfrac{\text{cos θ}}{\text{sin θ}} - \dfrac{1}{\text{sin θ}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos θ + 1 - sin θ}}{\text{sin θ}}}{\dfrac{\text{cos θ - 1 + sin θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + 1 - sin θ}}}{\text{cos θ - 1 + sin θ}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \\[1em] \Rightarrow \dfrac{{\text{cos θ + (1 - sin θ)}}}{\text{cos θ - (1 - sin θ)}} \times \dfrac{{\text{cos θ + (1 - sin θ)}}}{{\text{cos θ + (1 - sin θ)}}} \\[1em] \Rightarrow \dfrac{[\text{cos θ + (1 - sin θ)}]^2}{\text{cos}^2 θ - (\text{1 - sin θ})^2} \\[1em] \Rightarrow \dfrac{\text{cos}^2 \text{ θ} + \text{(1 - sin θ)}^2 + 2\text{cos θ(1 - sin θ)}}{\text{cos}^2 \text{ θ} - (1 + \text{sin}^2 θ - \text{2sin θ})} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ + \text{sin}^2 θ + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{\text{cos}^2 \text{ θ} - 1 - \text{sin}^2 θ + \text{2sin θ}} \\[1em] \Rightarrow \dfrac{1 + 1 + \text{2cos θ - 2sin θ - 2 sin θ cos θ}}{1 - \text{sin}^2 θ - 1 - \text{sin}^2 θ + \text{2 sin θ}} \\[1em] \Rightarrow \dfrac{2 + \text{2cos θ - 2sin θ - 2sin θ cos θ}}{\text{2sin θ - 2sin}^2 θ} \\[1em] \Rightarrow \dfrac{\text{2(1 + cos θ) - 2sin θ(1 + cos θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{(1 + cos θ)(2 - 2sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{2(1 + cos θ)(1 - sin θ)}}{\text{2sin θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{1 + cos θ}}{\text{sin θ}}.]

Since, L.H.S. = R.H.S. hence proved that cot θ + cosec θ - 1cot θ - cosec θ + 1=1 + cos θsin θ\dfrac{\text{cot θ + cosec θ - 1}}{\text{cot θ - cosec θ + 1}} = \dfrac{\text{1 + cos θ}}{\text{sin θ}}.

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