Is sin(x)-1/(sin(x))=-cos(x)cot(x) ?

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

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prove sin(x)-1/(sin(x))=-cos(x)cot(x)

prove

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

True

Solution steps

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

Manipulating right side

- cos (x) cot (x)

Express with sin, cos

- cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Remove parentheses:

(a) = a

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a, - (- a) = a

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

Cancel

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

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

Cancel the common factor:

sin (x)

= sin (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

2 cos^{2} (x) - cos (x) - 1 = 0

cos (θ) = cot (θ) sin^{2} (θ) csc (θ)

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

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)

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