Question

1+sin2a/1-sin2a=(1+tana/1-tana)^2

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Answer 1

Step-by-step explanation:

1+sin2a/1-sin2a=(1+tana/1-tana)^2

taking RHS

[(1+ tanA)/(1-tanA)]²

[(1+sinA/cosA)/(1-sinA/cosA)]²

[{(cosA+ sinA)/cosA}/{(cosA+ sinA)/cosA}]²

{cosA+sinA}²/{cosA-sinA}²

( sin²A+cos²A+2sinAcosA)/( sin²A+cos²A+2sinAcosA)

( 1+2sinAcosA)/(1-2sinAcosA)

2sinAcosA= sin2A

(1+sin2A)/(1-sin2A)

Answer 2

Step-by-step explanation:

cos2A /1-sin2A = 1+tanA/1-tanA

Note

1. cos2A= cos (square) A + sin ( square) A
2. 1= cos(square) theta + sin (square) theta.
3. sin2A =2cosA.sinA.

cos(square) A + sin (square) A / cos (square )A - 2cosA.sinA = cosA+ sin A/ cosA - sin A [ divid by cosA]

cosA / cosA + sinA / cosA / cosA / cosA - sinA / cosA = 1+ sinA/ cosA / 1- sinA / cosA

Answer : 1+ tanA/1- tanA = R.H.S