Question

(1-sinx+cosx)^2 = 2(1+sinx)(1+cosx)​

Answer 1

If you're trying to establish an identity, the given equation is not an identity. The proper identity would be as follows:

(1 - sin(x) + cos(x))² = (1 - sin(x))² + 2 (1 - sin(x)) cos(x) + cos²(x)

… = (1 - 2 sin(x) + sin²(x)) + 2 (1 - sin(x)) cos(x) + cos²(x)

… = 2 - 2 sin(x) + 2 (1 - sin(x)) cos(x)

… = 2 - 2 sin(x) + 2 cos(x) - 2 sin(x) cos(x)

… = 2 (1 - sin(x) + cos(x) - sin(x) cos(x))

… = 2 (1 - sin(x) + cos(x) (1 - sin(x)))

… = 2 (1 - sin(x)) (1 + cos(x))

But if you're trying to solve an equation:

(1 - sin(x) + cos(x))² = 2 (1 + sin(x)) (1 + cos(x))

2 (1 - sin(x)) (1 + cos(x)) = 2 (1 + sin(x)) (1 + cos(x))

(1 - sin(x)) (1 + cos(x)) - (1 + sin(x)) (1 + cos(x)) = 0

(1 + cos(x)) (1 - sin(x) - 1 - sin(x)) = 0

-2 sin(x) (1 + cos(x)) = 0

sin(x) = 0   or   1 + cos(x) = 0

sin(x) = 0   or   cos(x) = -1

[x = arcsin(0) + 2nπ   or   x = arcsin(0) + π + 2nπ]   or

… [x = arccos(-1) + 2nπ]

We have arcsin(0) = 0 and arccos(-1) = π, so the solution set reduces to

x = 2nπ   or   x = (2n + 1)π

(where n is any integer)