Solve identity,

(1-sin2x)/cos2x = cos2x/(1+sin2x)

I tried starting from the right side,
RS:
=(cos²x-sin²x)/(1+2sinxcosx)
=(cos²x-(1-cos²x))/(1+2sinxcosx)

and the right side just goes in circle. May I get a hint to start off?

First of all since the angle is 2x throughout, let's just use y for 2x

Secondly, you probably want to prove it as an identity, rather than solve it

RS
= cosy/(1+siny) [(1-siny)/(1-siny)]
= cosy(1-siny)/(1- sin^2 y)
= cosy(1-siny)/cos^2y
= (1-siny)/cosy)
= (1- sin 2x)/cos 2x
= LS

Cross multiply.

cos^2(2x) = 1- sin^2(2x) = cos^2 2x
q.e.d.

x can be anything.

That is why it is called an identity

Got it, thanks!

To solve the given identity, it is helpful to express the right side in terms of sine and cosine functions. We can apply the trigonometric identity: cos^2(x) = 1 - sin^2(x).

Starting from the right side:
R.S. = (cos^2(x))/(1 + sin^2(x))
= (1 - sin^2(x))/(1 + sin^2(x))

Now, we can try to manipulate the expression to achieve the form on the left side of the equation. Let's work on simplifying the denominator.

R.S. = (1 - sin^2(x))/(1 + sin^2(x))
= (1 - sin^2(x))/(1 + sin^2(x)) * (1 - sin^2(x))/(1 - sin^2(x)) [multiplying the numerator and the denominator by 1 - sin^2(x)]
= (1 - 2sin^2(x) + sin^4(x))/(1 - sin^2(x) + sin^2(x) - sin^4(x)) [using the distributive property]

Simplifying the numerator further, we can use the identity: sin^2(x) = 1 - cos^2(x).

R.S. = (1 - 2(1 - cos^2(x)) + (1 - cos^2(x))^2)/(1 - (1 - cos^2(x)) + (1 - cos^2(x)) - (1 - cos^2(x))^2)
= (1 - 2 + 2cos^2(x) + 1 - 2cos^2(x) + cos^4(x))/(1 - 1 + cos^2(x) - cos^2(x) + 1 - cos^2(x) + cos^4(x))

Now, simplify the expression.

R.S. = (2cos^2(x) + cos^4(x))/cos^4(x)
= cos^4(x)/cos^4(x) + 2cos^2(x)/cos^4(x)
= 1 + 2/cos^2(x)

Now, let's compare this with the left side of the equation: (1 - sin^2(x))/cos^2(x).

By applying the trigonometric identity: sin^2(x) = 1 - cos^2(x), we can simplify the left side of the equation.

L.S. = (1 - (1 - cos^2(x)))/cos^2(x)
= (1 - 1 + cos^2(x))/cos^2(x)
= cos^2(x)/cos^2(x)
= 1

Therefore, the left side (L.S.) is equal to the right side (R.S.), since both evaluate to 1.

Hence, the identity is verified and the solution is confirmed.