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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (- x) - sin (- x)

Use the negative angle identity:

sin (- x) = - sin (x)

= csc (- x) - (- sin (x))

Use the negative angle identity:

csc (- x) = - csc (x)

= - (- sin (x)) - csc (x)

Express with sin, cos

- (- sin (x)) - csc (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

- (- a) = a

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

Convert element to fraction:

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

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

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

sin (x) sin (x) - 1

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) - 1

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

- cos (x) cot (x)

- cos (x) cot (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (\frac{4 π}{3})

cot (x) = 1, 0 \leq x \leq 2 π

sin^{2} (x) = \frac{1}{4}

tan (- \frac{π}{4})

3 tan (θ) + 1 = 0

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