Question

help me with this trig $\bold\blue{\boxed{Question}}$ ​

Answer 1

Answer:

Please refer to the attachment

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

Prove That :- (1 + cosA + sinA) / (1 + cosA - sinA) = (1 + sinA)/cosA

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS Part :-

➻ (1 + cosA + sinA) / (1 + cosA - sinA)

Multiply & Divide By (1 + cosA + sinA) we get

➻ [(1 + cosA + sinA)(1 + cosA + sinA)] /[(1 + cosA - sinA)(1 + cosA + sinA)]

➻ [(1 + cosA + sinA)²] / [{(1 + cosA) - sinA} * {(1 + cosA) + sinA}]

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

➻ [(1 + cosA + sinA)²] / [ (1 + cosA)² - sin²A ]

using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca in Numerator Now,

➻ [ 1+cos²A+sin²A+2cosA+2sinAcosA+2sinA ] /[ (1 + cosA)² - sin²A ]

using sin² + cos²A = 1 in Numerator Now,

➻ [ 1 + 1 + 2sinA + 2cosA + 2sinAcosA ] / [(1 + cosA)² - sin²A ]

using (a + b)² = a² + b² + 2ab in Denominator now, and taking 2 common From Numerator

➻ 2[ 1 + sinA + cosA + sinAcosA ] / [ (1 + cos²A + 2cosA) - sin²A ]

Putting sin²A = 1 - cos²A in Denominator Now,

➻ 2[ (1 + cosA) + (sinA + sinAcosA) ] / [ ((1 + cos²A + 2cosA) - ( 1 - cos²A) ]

➻ 2[ (1 + cosA) + sinA(1 + cosA) ] / [ 1 + cos²A + 2cosA - 1 + cos²A ]

➻ 2(1+cosA)(1+sinA) / [ 2cos²A + 2cosA ]

Taking 2cosA common From Denominator Now,

➻ 2(1+cosA)(1+sinA) / 2cosA( 1 + cosA)

Now, 2cosA will be cancel , and, we get,

➻ (1 + sinA) / cosA = RHS . (Hence, Proved).

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