Question

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sinA / 1-cos A = 1+ cosA / sinA ​

Answer 1

Answer:

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Answer 2

Answer:

LHS = RHS.

Step-by-step explanation:

ATQ,

$\sf LHS = \dfrac{sinA}{1 - cosA}\\ \\ \\\sf Multiply \ the \ numerator \ and \ denominator \ by \ (1+cosA).\\ \\ \\\sf LHS = \dfrac{sinA}{1 - cosA} \ \ \dfrac{(1 + cosA)}{(1 + cosA)}\\ \\ \\ \\\sf LHS = \dfrac{sinA \ (1 + cosA)}{1 - cosA \ (1 + cosA)}\\ \\ \\ \\\boxed{ \ \sf Using \ the \ identity \ (a - b) \ (a + b) = a^2 - b^2 \ }\\ \\ \\ \\\sf LHS = \dfrac{sinA + (sinA\times cosA)}{1^2 - cos^2A}$

$\boxed{\sf Using \ 1^2 - cos^2A = sin^2A}\\ \\ \\ \\\sf LHS = \dfrac{sinA + (sinA \times cosA)}{sin^2A}\\ \\ \\ \\\sf LHS = \dfrac{sinA \left(1 + cosA\right)}{sin^2A}\\ \\ \\ \\\sf LHS = \dfrac{1 + cosA}{sinA}\\ \\ \\ \\\sf LHS = RHS\\ \\ \\ \\\sf Hence \ proved.$