Question

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

Answer 1

ㅤ$\;\;\underline{\textbf{\textsf{ To Prove :-}}}$

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

ㅤ$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

L.H.S

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

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

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

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

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

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

R.H.S

ㅤ$\;\;\underline{\textbf{\textsf{Hence -}}}$

$\leadsto \ {\boxed{\tt{LHS = RHS}}}$

$\therefore{ \underline{\bf{(Proved)}}}$

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