Question

18
If cos 0 + sin 0 = 1, prove that cos 0 -sin 0 = 1.
We have​

Answer 1

$cos\theta + sin\theta = 1$

$(cos\theta+sin\theta)^2 = 1$

$cos^2\theta+sin^2\theta + 2sin\theta cos\theta = 1$

$1 + 2sin\theta cos\theta = 1$

$2sin\theta cos\theta = 0\rightarrow(i)$

$Let, x= cos\theta - sin\theta$

$x^2 = (cos\theta - sin\theta)^2$

$x^2 = cos^2\theta + sin^2\theta - 2sin\theta cos\theta$

$x^2 = 1-2sin\theta cos\theta$

Substituting the value of 2sinθcosθ from equation (i), we get

$x^2 = 1$

$x = ^+_-1$

Hence cosθ - sinθ = ±1 (since x=±1), depending on the value of θ

Since we want to prove it to be equal to 1, we can write

Cosθ - Sinθ = 1

HOPE IT HELPS!!