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Mathematics

Prove the following identity:

(sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+cosec⁡θ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta

Trigonometric Identities

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Answer

Given equation,

⇒ (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Solving L.H.S. of the equation :

⇒ (sin θ + cos θ)(tan θ + cot θ)

⇒(sin⁡θ+cos⁡θ)(sin⁡θcos⁡θ+cos⁡θsin⁡θ)⇒(sin⁡θ+cos⁡θ)(sin⁡2θ+cos⁡2θcos⁡θ(sin⁡θ))⇒(sin⁡θ+cos⁡θ)(1cos⁡θsin⁡θ)⇒sin⁡θcos⁡θsin⁡θ+cos⁡θcos⁡θsin⁡θ⇒1cos⁡θ+1sin⁡θ⇒sec⁡θ+cosec⁡θ.\Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta}\Big) \\[1em] \Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta(\sin \theta)} \Big) \\[1em] \Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{1}{\cos \theta \sin \theta} \Big) \\[1em] \Rightarrow \dfrac{\sin \theta}{\cos \theta \sin \theta} + \dfrac{\cos \theta}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{1}{\cos \theta} + \dfrac{1}{\sin \theta} \\[1em] \Rightarrow \sec \theta + \cosec \theta.

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

Hence, proved that (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+cosec⁡θ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta.

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