Question

please solve for me​

Answer 1

Answer:

SORRY

Step-by-step explanation:

It is too hard .....

Answer 2

Step-by-step explanation:

This is easy my dear. Here's a proof.

SinA/(1 + CosA) + (1 + CosA)/SinA

= [(SinA)^2 + (1+CosA)^2]/[SinA(1 + CosA)]

= [(SinA)^2 + (CosA)^2 + 1 + 2CosA]/[SinA(1 + CosA)]

= [2(1 + CosA)]/[SinA(1 + CosA)] ( since, (SinA)^2 + (CosA)^2 = 1 )

= 2/SinA = 2CosecA, Hence LHS=RHS.

2. To prove the identity : [(1-cos A)/(1+cos A)]^0.5 = cosec A - cot A.

LHS = [(1-cos A)/(1+cos A)]^0.5

= [(1-cos^2 A/2+sin ^2 A/2)/(1+cos^2 A/2 - sin^2 A/2)]^0.5

= [(2 sin^2 A/2)/2 cos^2 A/2)]^0.5

= [tan^2 A/2]^0.5 = tan A/2 …(1)

RHS = cosec A - cot A

= (1/sin A)-(cos A/sin A)

= (1-cos A)/sin A

= (1-cos^2 A/2+sin^2 A/2)/2 sin A/2 cos A/2

= 2 sin^2 A/2/2 sin A/2 cos A/2

= (sin A/2)/(cos A/2)

= tan A/2 …(2)

Since (1) and (2) are the same the identity is proved.

Hope you understand.

Best wishes.

pls mark a s brainliest