prove thât sin A + cosA/sin A -cosA + sinA-cosA/sin A +cosA = 2/1-2cos^2 A

Let's delve into this trigonometric identity and prove the equation step by step. The expression we need to prove is:

Prove: (sin A + cos A) / (sin A - cos A) + (sin A - cos A) / (sin A + cos A) = 2 / (1 - 2cos² A)

Step 1: Simplifying the Left Side

To start, let's combine the two fractions on the left-hand side. The common denominator will be (sin A - cos A)(sin A + cos A). Thus, we rewrite the left side as follows:

• (sin A + cos A)² / [(sin A - cos A)(sin A + cos A)] + (sin A - cos A)² / [(sin A - cos A)(sin A + cos A)]

Now we can combine the numerators:

• Numerator: (sin A + cos A)² + (sin A - cos A)²

Step 2: Expanding the Numerators

Using the binomial expansion, we get:

• (sin² A + 2sin A cos A + cos² A) + (sin² A - 2sin A cos A + cos² A)

Combining these terms yields:

• 2sin² A + 2cos² A

Since sin² A + cos² A = 1, we can further simplify:

• 2(sin² A + cos² A) = 2(1) = 2

Step 3: Finalizing the Left Side

Now, substituting back into our fraction, we have:

• 2 / [(sin A - cos A)(sin A + cos A)]

Next, we simplify the denominator:

• (sin A)² - (cos A)² = sin² A - cos² A = (1 - cos² A) - cos² A = 1 - 2cos² A

So, we can express the left side of our equation as:

• 2 / (1 - 2cos² A)

Step 4: The Final Equality

At this point, we have shown that:

• (sin A + cos A) / (sin A - cos A) + (sin A - cos A) / (sin A + cos A) = 2 / (1 - 2cos² A)

This matches the right side of the original equation we set out to prove. Therefore, we've successfully demonstrated the identity!

Summary

By methodically simplifying and combining terms, we arrived at the conclusion that both sides of the equation are equal. This exercise not only proves the identity but also reinforces the fundamental relationships in trigonometry, particularly the Pythagorean identity. If you have any further questions about trigonometric identities or need clarification on any steps, feel free to ask!