Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Given equation,

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

Solving L.H.S. of the equation :

$\Rightarrow \text{(sin θ + cos θ)(tan θ + cot θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{\text{cos θ}}{\text{sin θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin}^2 θ + \text{cos}^2 θ}{\text{cos θ sin θ}}\Big)$

By formula,

sin² θ + cos² θ = 1.

$\Rightarrow \text{(sin θ + cos θ)} \times \dfrac{1}{\text{cos θ sin θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ}}{\text{cos θ sin θ}} + \dfrac{\text{cos θ}}{\text{cos θ sin θ}} \\[1em] \Rightarrow \dfrac{1}{\text{cos θ}} + \dfrac{1}{\text{sin θ}} \\[1em] \Rightarrow \text{sec θ + cosec θ}.$

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