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Mathematics

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

tan2 θtan2 θ−1+cosec2 θsec2 θ−cosec2 θ=1sin2 θ−cos2 θ\dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}.

Trigonometric Identities

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Answer

Solving L.H.S.,

⇒sin2 θcos2 θsin2 θcos2 θ−1+1sin2 θ1cos2 θ−1sin2 θ=sin2 θcos2 θsin2 θ−cos2 θcos2 θ+1sin2 θsin2 θ−cos2 θcos2 θ sin2 θ=sin2 θsin2 θ−cos2 θ+1sin2 θ−cos2 θcos2 θ=sin2 θsin2 θ−cos2 θ+cos2 θsin2 θ−cos2 θ=sin2 θ+cos2 θsin2 θ−cos2 θ=1sin2 θ−cos2 θ.\Rightarrow \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}} - 1} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{1}{\text{cos}^2 \text{ θ}} - \dfrac{1}{\text{sin}^2 \text{ θ}}} \\[1em] = \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} \text{ sin}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{1}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}.

Since, L.H.S. = R.H.S. hence proved that,

tan2 θtan2 θ−1+cosec2 θsec2 θ−cosec2 θ=1sin2 θ−cos2 θ\dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}.

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