Question

pls answer...it's urgent​

Answer 1

Answer :-

→ 2 sinA / (1 + sin A + cos A)

multiply numerator and denominator by (1 + sin A - cos A)

→ {2 sinA / (1 + sin A + cos A)} / {(1 + sin A - cos A) / (1 + sin A - cos A)}

→ 2sinA(1 + sin A - cos A) / (1 + sin A + cos A)(1 + sin A - cos A)

using (a + b)(a - b) = a² - b² in denominator,

→ 2sinA(1 + sin A - cos A) / (1 + sinA)² - cos²A

using cos²A = 1 - sin²A in denominator,

→ 2sinA(1 + sin A - cos A) / (1 + sinA)² - (1 - sin²A)

→ 2sinA(1 + sin A - cos A) / (1 + sin²A + 2sinA - 1 + sin²A)

→ 2sinA(1 + sin A - cos A) / (2sinA + 2sin²A)

→ 2sinA(1 + sin A - cos A) 2sinA( 1 + sinA)

→ (1 + sin A - cos A) / (1 + sinA)

now ,given that,

→ 2 sinA / (1 + sin A + cos A) = k

so,

→ (1 + sin A - cos A) / (1 + sinA) = k

also, given,

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

therefore,

→ k = 2 (Ans.)

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