Prove that :

sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A

To prove:

sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A

Solving L.H.S. of the equation :

$\Rightarrow \text{sin A(1 + tan A) + cos A(1 + cot A)} \\[1em] \Rightarrow \text{sin A + sin A tan A + cos A + cos A cot A} \\[1em] \Rightarrow \text{sin A + cos A} + \text{sin A} \times \dfrac{\text{sin A}}{\text{cos A}} + \text{cos A} \times \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + cos A} + \dfrac{\text{sin}^2 A}{\text{cos A}} + \dfrac{\text{cos}^2 A}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + cos A} + \dfrac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A cos A}} \\[1em] \Rightarrow \text{sin A + cos A} + \dfrac{\text{(sin A + cos A)}(\text{sin}^2 A + \text{cos}^2 A - \text{sin A cos A})}{\text{sin A cos A}} \\[1em] \Rightarrow \text{(sin A + cos A)}\Big(1 + \dfrac{\text{sin}^2 A + \text{cos}^2 A - \text{sin A cos A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow \text{(sin A + cos A)}\Big(1 + \dfrac{1 - \text{sin A cos A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow \text{(sin A + cos A)}\Big(\dfrac{\text{sin A cos A} + 1 - \text{sin A cos A}}{\text{sin A cos A}}\Big) \\[1em] \Rightarrow \text{(sin A + cos A)} \times \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A + cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{sin A cos A}} + \dfrac{\text{cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow \text{sec A + cosec A}.$

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

Hence, proved that sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A.