Question

prove sinA/1+cosA=1-cosA/sinA​

Answer 1

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Answer 2

To prove:

$\dfrac{sinA}{1+cosA} = \dfrac{1-cosA}{sinA}$

Proof:

LHS = $\dfrac{sinA}{1+cosA}$

Rationalising the denominator

$= \dfrac{sinA}{1-cosA} \times \dfrac{1-cosA}{1-cosA}$

$= \dfrac{sinA(1-cosA)}{(1+cosA)(1-cosA)}$

$= \dfrac{sinA(1-cosA)}{1-{cos}^{2}A}$

$= \dfrac {sinA(1-cosA)}{{sin}^{2}A}$ $=\dfrac{1-cosA}{sinA} = RHS$

Identities used

• $({a}^{2} - {b}^{2}) = (a+b)(a-b)$
• ${sin}^{2}\theta + {cos}^{2}\theta = 1$

Hence, proved.