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

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

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

prove

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

True

Solution steps

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

Use the basic trigonometric identity:

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

\frac{cos ^{2} ( x )}{\frac{1}{c s c ( x )}}

Simplify

\frac{cos ^{2} ( x )}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

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

= \frac{cos ^{2} ( x ) csc ( x )}{1}

Apply rule

\frac{a}{1} = a

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

sin (x) cot (x) = cos (x)

sin (π + x) = - sin (x)

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

1 + tan^{2} (α) = \frac{1}{cos ^{2} ( α )}

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