Math, asked by adev141203, 1 year ago

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

Answers

Answered by kishan9163
8
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Answered by Tomboyish44
17

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.

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