We are asked to prove the trigonometric identity $(1 + tan A)^2 + (1 + cot A)^2 = (sec A + csc A)^2$.

1. Problem Description

We are asked to prove the trigonometric identity $(1 + tan A)^2 + (1 + cot A)^2 = (sec A + csc A)^2$.

2. Solution Steps

We will start by expanding the left-hand side (LHS) of the equation:

$(1 + tan A)^2 + (1 + cot A)^2 = (1 + 2 tan A + tan^2 A) + (1 + 2 cot A + cot^2 A)$

$= 2 + 2 tan A + 2 cot A + tan^2 A + cot^2 A$

We know the following trigonometric identities:

$tan A = \frac{sin A}{cos A}$

$cot A = \frac{cos A}{sin A}$

$sec A = \frac{1}{cos A}$

$csc A = \frac{1}{sin A}$

$sin^2 A + cos^2 A = 1$

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

$cot^2 A + 1 = csc^2 A$

From the Pythagorean identities, we can replace $tan^2 A$ and $cot^2 A$ in our expression:

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

$cot^2 A = csc^2 A - 1$

Substituting these into our expression, we get:

$2 + 2 tan A + 2 cot A + tan^2 A + cot^2 A = 2 + 2 tan A + 2 cot A + (sec^2 A - 1) + (csc^2 A - 1)$

$= 2 + 2 tan A + 2 cot A + sec^2 A - 1 + csc^2 A - 1$

$= 2 tan A + 2 cot A + sec^2 A + csc^2 A$

Now, let's expand the right-hand side (RHS) of the equation:

$(sec A + csc A)^2 = sec^2 A + 2 sec A csc A + csc^2 A$

Now, we need to show that $2 tan A + 2 cot A = 2 sec A csc A$.

Let's rewrite $2 tan A + 2 cot A$:

$2 tan A + 2 cot A = 2 \frac{sin A}{cos A} + 2 \frac{cos A}{sin A} = 2 (\frac{sin^2 A + cos^2 A}{sin A cos A}) = 2 (\frac{1}{sin A cos A})$

Since $sec A = \frac{1}{cos A}$ and $csc A = \frac{1}{sin A}$, then $sec A csc A = \frac{1}{cos A sin A}$.

Therefore, $2 sec A csc A = \frac{2}{sin A cos A}$.

So, $2 tan A + 2 cot A = 2 sec A csc A$.

Thus, the left-hand side is $2 tan A + 2 cot A + sec^2 A + csc^2 A = 2 sec A csc A + sec^2 A + csc^2 A$.

And the right-hand side is $sec^2 A + 2 sec A csc A + csc^2 A$.

Therefore, $(1 + tan A)^2 + (1 + cot A)^2 = (sec A + csc A)^2$ is true.

3. Final Answer

$(1 + tan A)^2 + (1 + cot A)^2 = (sec A + csc A)^2$