Question

(sinA + cosA) / (sinA - cosA) + (sinA - cosA) / (sinA + cosA) = 2 / sin^2A - cos^2A = 2 / 2sin^A -1

Answer 1

this is solution please open image and see the solution

Answer 2

To prove:

$\implies \sf \dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA} = \dfrac{2}{sin^2A - cos^2A} = \dfrac{2}{2sin^2A - 1}$

LHS:

$\implies \sf \dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA}$

$\implies \sf \dfrac{sinA + cosA \big(sinA + cosA\big) + sinA - cosA\big(sinA - cosA\big)}{\big(sinA - cosA\big)\big(sinA + cosA\big)}$

Apply these identities:

→ (a + b)² = a² + b² + 2ab

→ (a - b)² = a² + b² - 2ab

→ a² - b² = (a + b)(a - b)

$\implies \sf \dfrac{\big(sinA + cosA\big)^2 + \big(sinA - cosA\big)^2}{sin^2A - cos^2A}$

$\implies \sf \dfrac{sin^2A + cos^2A + 2(sinA)(cosA) + sin^2A + cos^2A - 2(sinA)(cosA)}{sin^2A - cos^2A}$

Using sin²A + cos²A = 1 & after cancelling 2 sinA cosA and -2 sinA cosA we get:

$\implies \sf \dfrac{1 + 1}{sin^2A - cos^2A}$

$\implies \sf \dfrac{2}{sin^2A - cos^2A}$

Part 1 proved.

Now, substitute cos²A = 1 - sin²A

$\implies \sf \dfrac{2}{sin^2A - \big[1 - sin^{2}A\big]}$

$\implies \sf \dfrac{2}{sin^2A - 1 + sin^{2}A}$

$\implies \sf \dfrac{2}{2sin^2A - 1}$

Part 2 proved.

Therefore, LHS = RHS.