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

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

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

prove

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

True

Solution steps

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

Manipulating left side

(cot (x) - csc (x)) (cos (x) + 1)

Express with sin, cos

(1 + cos (x)) (cot (x) - csc (x))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

(1 + cos (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 )} (cos (x) + 1)

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

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

Apply Difference of Two Squares Formula:

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

a = cos (x), b = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= cos^{2} (x) - 1

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

Use the Pythagorean identity:

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

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

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

Simplify

\frac{- sin ^{2} ( x )}{sin ( x )} : - sin (x)

\frac{- sin ^{2} ( x )}{sin ( x )}

Apply the fraction rule:

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

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

Cancel the common factor:

sin (x)

= - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 csc^{2} (θ) - 1

p ro v e \frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

p ro v e csc (θ) - sin (θ) = cos (θ) cot (θ)

p ro v e sec (x) (1 - sin^{2} (x)) = cos (x)

p ro v e (1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1

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