Question

cosA/1+sinA + 1+sinA/cosA =2sinA​

Answer 1

Answer:

first of all your question is wrong.

the correct question is cosA/1+sinA + 1+sinA/cosA = 2secA

Step-by-step explanation:

cos²A+(1+sinA)²/cos(1+sinA)  (by cross multiplying for lcm)

cos²A + 1² + sin²A + 2sinA/cosA(1+sinA)

when cos²A + sin²A = 1, then

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

2+2sinA/cosA(1+sinA)

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

2/cosA

= 2 secA

hope it helps you

please mark brainliest :)

Answer 2

Answer:

The answer is 2secA and not 2sin A

Step-by-step explanation:

Step 1: Multiplying to make the denominators equal

cosA/(1+sinA) + (1+sinA)/cosA  = cosA(cosA/1+sinA) + 1+sinA( 1+SinA /cosA )

Step 2: Solving we get

=($cos^{2}A + sin^2{A$ + 1 + sinA + sinA) / (1+sinA)(cosA)

Since cos^{2}A + sin^2{A} = 1 we get

=(2+2sinA) / (cosA + sinAcosA)

Step 3: Taking out the common terms from numerator and denominator we get,

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

Cancelling out the common term (1+sinA) we get 2/cosA

= 2(1/cosA)

=2secA