Question

1-tana/1+tana=1-sin2a/cos2a

Answer 1

Answer:

Hey mate..........

Step-by-step explanation:

(1-tan A)/(1+tanA)= (cos 2A)/(1+sin 2A)

To prove that the LHS (left hand side) = RHS (right hand side) we must manipulate the expressions. Changing the left hand side

1. 1-tan A =1-sinA/cosA (numerator)

2. 1+tan A = 1+sinA/cosA(denominator)

Now use the LCD for 1. and 2. which iscosA

1. therefore 1-sinA/cosA = (cosA - sinA)/cos A (numerator) and

2.therefore 1+sinA/cosA = (cosA + sinA)/cos A (denominator)

Now 1. is divided by 2.

therefore ((cosA-sinA)/cosA) divide ((cosA+sinA)/cosA)

therefore =(cosA-sinA)/(cosA+sinA)because the cos As will cancel out (divide )

Now multiply our new expression by(cosA +sin A)/(cosA+sinA)because it will allow us to simplify effectively to obtain the RHS and is the same as1/1.

therefore = (cosA-sinA)/(cosA+sinA) times (cosA+sinA)/(cosA+sinA)

therefore = (cos^2A - sin^2A)/(cosA + sinA)^2

We know from double angle identities that :cos^2A-sin^2=cos2A

We know from the perfect square (our denominator) that:

(cosA+sinA)^2 = cos^2A +2sinAcosA + sin^2A

Now rearrange the order of the square:

cos^2A +sin^2A + 2sinAcosA

We know from identities thatcos^2A +sin^2A = 1

and2sinAcosA = sin2A

Now put all the information of our denominator together:

therefore (cosA+sinA)^2 = 1+sin2A

Now combine with the numerator which we recall as:

sin^2A-cos^2a = cos 2A

therefore = (cos2A)/ (1+sin2A)

therefore LHS=RHS

Hope it helped......