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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cot (- x) cos (- x) + sin (- x)

Use the negative angle identity:

sin (- x) = - sin (x)

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Use the negative angle identity:

cot (- x) = - cot (x)

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

Express with sin, cos

- sin (x) + (- cot (x)) cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

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

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

Multiply fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

- \frac{1}{\frac{1}{c s c ( x )}}

Simplify

- \frac{1}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= - csc (x)

- csc (x)

- csc (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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