Prove that :

sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A

To prove:

Equation : sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A.

Solving L.H.S. of the above equation :

$\Rightarrow \text{sin}^2 A \text{ tan A + cos}^2 A \text{ cot A + 2 sin A cos A} \\[1em] \Rightarrow \text{sin}^2 A \times \dfrac{\text{sin A}}{\text{cos A}} + \text{cos}^2 A \times \dfrac{\text{cos A}}{\text{sin A}} + \text{2 sin A cos A} \\[1em] \Rightarrow \dfrac{\text{sin}^3 A}{\text{cos A}} + \dfrac{\text{cos}^3 A}{\text{sin A}} + \text{2 sin A cos A} \\[1em] \Rightarrow \dfrac{\text{sin}^4 A + \text{cos}^4 A + \text{2 sin}^2 A \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{(\text{sin}^2 A + \text{cos}^2 A)^2}{\text{sin A cos A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1}{\text{sin A cos A}}$.

Solving R.H.S. of the above equation :

$\Rightarrow \text{tan A + cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A cos A}}.$

Since, L.H.S. = R.H.S. = $\dfrac{1}{\text{sin A cos A}}$.

Hence, proved that sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A.