Is csc(θ)-cot(θ)=(1-cos(θ))/(sin(θ)) ?

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

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prove csc(θ)-cot(θ)=(1-cos(θ))/(sin(θ))

prove

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

True

Solution steps

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

Manipulating left side

csc (θ) - cot (θ)

Express with sin, cos

- cot (θ) + csc (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

- \frac{cos ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )} : \frac{- cos ( θ ) + 1}{sin ( θ )}

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

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- cos ( θ ) + 1}{sin ( θ )}

= \frac{- cos ( θ ) + 1}{sin ( θ )}

= \frac{1 - cos ( θ )}{sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (x) = tan (x) + 2 sin^{2} (x)

p ro v e \frac{1}{2 cot ( 1 - cos ^{2} ( x ) )} = csc (2 x)

p ro v e cot (u) = \frac{1}{tan ( u )}

p ro v e - sin (2 x) = - 4 cos (x) sin (x)

p ro v e cos (θ) (tan (θ) - sec (- θ)) = sin (θ) - 1

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