Is (sin(θ)-cos(θ))^2-1=-2sin(θ)cos(θ) ?

The answer to whether (sin(θ)-cos(θ))^2-1=-2sin(θ)cos(θ) is True

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prove (sin(θ)-cos(θ))^2-1=-2sin(θ)cos(θ)

prove

(sin (θ) - cos (θ))^{2} - 1 = - 2 sin (θ) cos (θ)

True

Solution steps

(sin (θ) - cos (θ))^{2} - 1 = - 2 sin (θ) cos (θ)

Manipulating left side

(sin (θ) - cos (θ))^{2} - 1

Rewrite using trig identities

= 2 sin^{2} (θ) - 2 sin (θ) cos (θ)

= - 2 sin (θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (2 π) = tan (2 π)

sec (θ) cot (θ) - sin (θ) = cos (θ) cot (θ)

9 sin (9 π - x) = 9 sin (x)

csc (u) - cos (u) = \frac{sin ( u )}{1 + cos ( u )}

sec (x) \cdot cot (x) = csc (x)

Is (sin(θ)-cos(θ))^2-1=-2sin(θ)cos(θ) ?
The answer to whether (sin(θ)-cos(θ))^2-1=-2sin(θ)cos(θ) is True