Is cos(x)-csc(x)cot(x)=-cos(x)cot^2(x) ?

The answer to whether cos(x)-csc(x)cot(x)=-cos(x)cot^2(x) is True

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Popular Trigonometry >

prove cos(x)-csc(x)cot(x)=-cos(x)cot^2(x)

Solution

prove

cos (x) - csc (x) cot (x) = - cos (x) cot^{2} (x)

Solution

True

Solution steps

cos (x) - csc (x) cot (x) = - cos (x) cot^{2} (x)

Manipulating left side

cos (x) - csc (x) cot (x)

Express with sin, cos

cos (x) - cot (x) csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply:

cos (x) \cdot 1 = cos (x)

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

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

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

Factor

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

- cos (x) + cos (x) sin^{2} (x)

Factor out common term

cos (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

\frac{( - cos ^{2} ( x ) ) cos ( x )}{sin ^{2} ( x )} : - \frac{cos ^{3} ( x )}{sin ^{2} ( x )}

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

Remove parentheses:

(- a) = - a

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

- cos^{2} (x) cos (x) = - cos^{3} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= - cos^{3} (x)

= \frac{- cos ^{3} ( x )}{sin ^{2} ( x )}

Apply the fraction rule:

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

= - \frac{cos ^{3} ( x )}{sin ^{2} ( x )}

= - \frac{cos ^{3} ( x )}{sin ^{2} ( x )}

= - \frac{cos ^{3} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

- cos (x) cot (x) cot (x)

Simplify

- cos (x) cot (x) cot (x) : - cot^{2} (x) cos (x)

- cos (x) cot (x) cot (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (x) cot (x) = cot^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

- cot^{2} (x) cos (x)

- cot^{2} (x) cos (x)

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(x)-csc(x)cot(x)=-cos(x)cot^2(x) ?
The answer to whether cos(x)-csc(x)cot(x)=-cos(x)cot^2(x) is True