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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

Rewrite using trig identities

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

Multiply by

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

= \frac{- sin ( x )}{\frac{( 1 - c o s ( x ) ) ( 1 + c o s ( x ) )}{1 + c o s ( x )}}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= \frac{- sin ( x )}{\frac{1 - c o s ^{2} ( x )}{1 + c o s ( x )}}

Use the Pythagorean identity:

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

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

= \frac{- sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Simplify

\frac{- sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}} : - \frac{1 + cos ( x )}{sin ( x )}

\frac{- sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Apply the fraction rule:

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

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}} = \frac{sin ( x ) ( 1 + cos ( x ) )}{sin ^{2} ( x )}

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

Cancel the common factor:

sin (x)

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a

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

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

Simplify

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

Manipulating right 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 )}

We showed that the two sides could take the same form

\Rightarrow True

(tan (b) + sec (b)) (1 - sin (b)) = cos (b)

cot (2 x) = \frac{cos ( 2 x )}{sin ( 2 x )}

cos^{2} (a) + cos^{2} (a) cot^{2} (a) = cot^{2} (a)

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

\frac{csc ( a )}{cot ( a )} = sec (a)

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