Prove the trigonometric identity $(1 + \tan A)^2 + (1 + \cot A)^2 = (\sec A + \csc A)^2$.

1. Problem Description

Prove the trigonometric identity $(1 + \tan A)^2 + (1 + \cot A)^2 = (\sec A + \csc A)^2$.

2. Solution Steps

We will expand both sides of the equation. Let's start with the left-hand side (LHS):

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

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

Now, let's work with the right-hand side (RHS):

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

We know that $\sec^2 A = 1 + \tan^2 A$ and $\csc^2 A = 1 + \cot^2 A$. Substituting these into the RHS expression, we get:

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

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

Recall that $\tan A = \frac{\sin A}{\cos A}$, $\cot A = \frac{\cos A}{\sin A}$, $\sec A = \frac{1}{\cos A}$, and $\csc A = \frac{1}{\sin A}$.

So, we have

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

Also,

$2\sec A \csc A = 2\frac{1}{\cos A} \frac{1}{\sin A} = 2\frac{1}{\sin A \cos A}$

Therefore, $2\tan A + 2\cot A = 2\sec A \csc A$.

Substituting this back into the LHS, we have:

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

Since LHS = RHS, the identity holds.

3. Final Answer

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