Is 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) ?

The answer to whether 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) is True

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prove 1/(sin(θ)cos(θ))-tan(θ)=cot(θ)

prove

\frac{1}{sin ( θ ) cos ( θ )} - tan (θ) = cot (θ)

True

Solution steps

\frac{1}{sin ( θ ) cos ( θ )} - tan (θ) = cot (θ)

Manipulating left side

\frac{1}{sin ( θ ) cos ( θ )} - tan (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

(- sin (x)) (3 cos (x)) = - 3 sin (x) cos (x)

cot (θ) \cdot sec (θ) = \frac{1}{cos ( θ )}

sin (x) sin (y) = 2 sin (\frac{x + y}{2}) cos (\frac{x - y}{2})

1 - sin (x) \cdot cos (x) \cdot tan (x) = cos^{2} (x)

sin (x) cos (x) = sec^{2} (x) - csc^{2} (x)

Is 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) ?
The answer to whether 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) is True