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Mathematics

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

cos θ1 - tan θ−sin2θcos θ - sin θ=cos θ + sin θ\dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} = \text{cos θ + sin θ}.

Trigonometric Identities

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Answer

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

⇒cos θ1 - tan θ−sin2θcos θ - sin θ⇒cos θ1−sin θcos θ−sin2θcos θ - sin θ⇒cos2θcos θ - sin θ−sin2θcos θ - sin θ⇒cos2θ−sin2θcos θ - sin θ⇒(cos θ - sin θ)(cos θ + sin θ)cos θ - sin θ⇒cos θ + sin θ.\Rightarrow \dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos θ}}{{1 - \dfrac{\text{sin θ}}{\text{cos θ}}}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{cos θ - sin θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \dfrac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ - sin θ}} \\[1em] \Rightarrow \text{cos θ + sin θ}.

Since, L.H.S. = R.H.S. hence, proved that cos θ1 - tan θ−sin2θcos θ - sin θ=cos θ + sin θ\dfrac{\text{cos θ}}{\text{1 - tan θ}} - \dfrac{\text{sin}^2 θ}{\text{cos θ - sin θ}} = \text{cos θ + sin θ}.

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