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

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

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

prove

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

True

Solution steps

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

Rewrite using trig identities

Multiply by

\frac{1 + cos ( θ )}{1 + cos ( θ )}

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

Expand

(1 + cos (θ)) (1 - cos (θ)) : 1 - cos^{2} (θ)

(1 + cos (θ)) (1 - cos (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{1 + cos ( θ )}

= \frac{sin ( θ )}{1 + cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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