Question

sinA/1+cosA + sinA/1-cosA= 2cosecA​

Answer 1

Answer:

Hence,proved.

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Answer 2

$\sf Question: To \ Prove \ \dfrac{sinA}{1 + cosA} + \dfrac{sinA}{1 - cosA} = 2 \ cosecA \:$

Proof:

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

• Taking sinA as common outside we get,

$\sf LHS = sinA \times \left(\dfrac{1}{1 + cosA} + \dfrac{1}{1 - cosA}\right)$

$\sf LHS = sinA \times \left( \ \dfrac{1 - cosA + 1 + cosA}{(1 + cosA)(1 - cosA)} \ \right)$

• Using (a - b)(a + b) = a² - b²

$\sf LHS = sinA \times \left(\dfrac{2}{(1^2 - cos^2A)}\right)$

• Using 1 - cos²A = sin²A we get,

$\sf LHS = sinA \times \left(\dfrac{2}{sin^2A}\right)$

• Cancelling sinA we get,

$\sf LHS = \left(\dfrac{2}{sinA}\right) \:$

$\sf LHS = 2 \times \left(\dfrac{1}{sinA}\right) \:$

• But 1/sinA = cosecA, therefore,

$\sf {LHS = 2 \times cosecA} \:$

$\sf {LHS = 2 \ cosecA}$

$\sf{LHS = RHS} \:$

Hence Proved.