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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (x) - cot (x)

Express with sin, cos

- cot (x) + csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

= \frac{1 - cos ( x )}{sin ( x )}

Rewrite using trig identities

Multiply by

\frac{1 + cos ( x )}{1 + cos ( x )}

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

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (x)

= 1^{2} - cos^{2} (x)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

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

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

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

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

Cancel the common factor:

sin (x)

= \frac{sin ( x )}{1 + cos ( x )}

= \frac{sin ( x )}{1 + cos ( x )}

We showed that the two sides could take the same form

\Rightarrow True

sec (2 x) = \frac{sec ^{2} ( x )}{2 - sec ^{2} ( x )}

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

(csc^{2} (x) - 1) sin^{2} (x) = cos^{2} (x)

cos (θ) - sec (θ) = - sin (θ) tan (θ)

tan (a) + cot (a) = sec (a) csc (a)

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