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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cot (- a) cos (- a) + sin (- a)

Use the negative angle identity:

sin (- x) = - sin (x)

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Use the negative angle identity:

cot (- x) = - cot (x)

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

Express with sin, cos

- sin (a) + (- cot (a)) cos (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

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

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

Multiply fractions:

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

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

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

cos (a) cos (a)

Apply exponent rule:

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

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

= cos^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cos^{2} (a)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

sin (a) sin (a)

Apply exponent rule:

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

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

= sin^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= sin^{2} (a)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= - csc (a)

- csc (a)

- csc (a)

We showed that the two sides could take the same form

\Rightarrow True

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