We are asked to prove the trigonometric identity: $sinA(cscA + sinAsec^2A) = sec^2A$

1. Problem Description

We are asked to prove the trigonometric identity:

$sinA(cscA + sinAsec^2A) = sec^2A$

2. Solution Steps

We will start with the left side of the equation and try to simplify it to match the right side.

$sinA(cscA + sinAsec^2A)$

First, distribute the $sinA$ term:

$sinA \cdot cscA + sinA \cdot sinA \cdot sec^2A$

Recall the definition of $cscA$ and $secA$:

$cscA = \frac{1}{sinA}$

$secA = \frac{1}{cosA}$

Substitute $cscA = \frac{1}{sinA}$ into the expression:

$sinA \cdot \frac{1}{sinA} + sin^2A \cdot sec^2A$

Simplify the first term:

$1 + sin^2A \cdot sec^2A$

Substitute $secA = \frac{1}{cosA}$ into the expression:

$1 + sin^2A \cdot \frac{1}{cos^2A}$

$1 + \frac{sin^2A}{cos^2A}$

Recall the definition of $tanA$:

$tanA = \frac{sinA}{cosA}$

Substitute $tanA = \frac{sinA}{cosA}$ into the expression:

$1 + tan^2A$

Recall the Pythagorean trigonometric identity:

$1 + tan^2A = sec^2A$

Substitute $1 + tan^2A = sec^2A$ into the expression:

$sec^2A$

This matches the right side of the original equation.

3. Final Answer

$sinA(cscA + sinAsec^2A) = sec^2A$

The identity is proven.