Is (csc^2(x)-1)sin(x)=cot(x)cos(x) ?

The answer to whether (csc^2(x)-1)sin(x)=cot(x)cos(x) is True

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prove (csc^2(x)-1)sin(x)=cot(x)cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

(- 1 + csc^{2} (x)) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= \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 )}

= \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)

= cot (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) (csc^{2} (x) - 1) = cot (x)

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

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

\frac{sec ^{2} ( t )}{tan ( t )} = sec (t) csc (t)

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

Is (csc^2(x)-1)sin(x)=cot(x)cos(x) ?
The answer to whether (csc^2(x)-1)sin(x)=cot(x)cos(x) is True