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Mathematics

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

(sin θ + cos θ)(sec θ + cosec θ) = 2 + sec θ cosec θ.

Trigonometric Identities

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Answer

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

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

Since, L.H.S. = R.H.S. hence proved that, (sin θ + cos θ)(sec θ + cosec θ) = 2 + sec θ cosec θ.

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