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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(cos (θ) - sin (θ))^{2}

Expand

(cos (θ) - sin (θ))^{2} : cos^{2} (θ) - 2 cos (θ) sin (θ) + sin^{2} (θ)

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

Rewrite using trig identities

= 1 - 2 cos (θ) sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

- cot (2 x) + csc (2 x) = tan (x)

sin (π + 0) = - sin (0)

\frac{csc ^{2} ( θ ) - 1}{cos ^{2} ( θ )} = csc^{2} (θ)

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

1 = csc^{2} (z) - cot^{2} (z)

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