# Prove the following identity (1+2(sin a)(cos a))/(1-2(sin^2 a)=(cos a+sin a)/(cos a-sin a)

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We have to prove that (1+2(sin a)(cos a))/(1-2(sin^2 a)=(cos a+sin a)/(cos a-sin a)

Starting from the left hand side

(1+2(sin a)(cos a))/(1-2(sin a)^2)

replace 1 with (sin a)^2 + (cos a)^2

=> [(sin a)^2 + (cos a)^2 + 2(sin a)(cos a)] / [(sin a)^2 + (cos a)^2 - 2(sin a)^2]

=> ( sin a + cos a)^2 / [(cos a)^2 - (sin a)^2]

=> ( sin a + cos a)^2 / [(cos a - sin a)(cos a + sin a)]

=> (sin a + cos a) / ( cos a - sin a)

which is the right side side

**We prove (1+2(sin a)(cos a))/(1-2(sin^2 a)=(cos a+sin a)/(cos a-sin a)**

We recognize the formula of the cosine of double angle for:

1-2(sin a)^2 = cos 2a

Also, we recognize the formula for the sine of double angle:

2sina*cosa = sin2a

We'll re-write the ratio from the left side:

(1 + sin 2a)/cos 2a =(cos a+sin a)/(cos a-sin a)

But cos 2a = (cos a)^2 - (sin a)^2

cos 2a = (cos a - sin a)(cos a + sin a)

We'll re-write the identity, substituting cos 2a:

(1 + sin 2a)/(cos a - sin a)(cos a + sin a) =(cos a+sin a)/(cos a-sin a)

We'll simplify and we'll get:

(1 + sin 2a)/(cos a + sin a) = (cos a+sin a)

1 + sin 2a = (cos a+sin a)^2

We'll expand the square:

1 + sin 2a = (cos a)^2+ 2sin a*cos a + (sin a)^2

But (cos a)^2 + (sin a)^2 = 1

1 + sin 2a = 1 + 2sin a*cos a

** 1 + sin 2a = 1 + sin 2a q.e.d.**

* R.H.S.*= (cos a+sin a)/(cos a-sin a)

=(cos a+sin a)/(cos a-sin a) x (cos a+sin a)/(cos a+sin a)

=(cos^2 a+2(cos a)(sin a)+sin^2 a)/(cos^2 a-sin^2 a)

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

=**L.H.S**