Question

I need to demostrate this trigonometric identities URGENT!

1. SinA-2sin3A / 2cos3A - cosA = Tan A

2. 1+ sin A / 1 - sin A - 1-Sin A / 1+ Sin A = 4 Tan A sec A

Answer 1

Step-by-step explanation:

(1).

SinA-2sin3A / 2cos3A - cosA

= sinA{1-2(sinA)^2}/cosA(2cos^2A - 1)

= sinA[(CosA)^2-(SinA)^2]/cosA(2cos^2A - 1)

= SinA(Cos2A)/cosA(2cos^2A - 1)

= TanA * Cos2A/[(cosA)^2 + cosA^2-CosA^2 - SinA^2]

[Because , Cos^A - SinA^2 = 1]

= TanA * Cos2A/[(cosA)^2 - SinA^2]

= TanA

(2).

1+ sin A / 1 - sin A - 1-Sin A / 1+ Sin A = 4 Tan A sec A

LHS

={(1+ sin A)/(1- sin A)}-(1-Sin A)/(1+Sin A)

= (1+sinA)^2/(cosA)^2 -(1-Sin A)^2/(cosA)^2

= (1/cosA)^2[4sinA]

= TanA SecA = RHS

Answer 2

Answer:

Hence Proved

Step-by-step explanation:

See the attachments