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

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

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

prove

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

True

Solution steps

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

Manipulating right 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

(cot (x) + tan (x))^{2} = csc^{2} (x) sec^{2} (x)

(sec (a)) (sin (a)) = tan (a)

cos (3 x) = 4 cos (3 x) - 3 cos (x)

\frac{1 + sec ^{2} ( x )}{2 + tan ^{2} ( x )} = 1

sec^{2} (a) + csc^{2} (a) = sec^{2} (a) csc^{2} (a)

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