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

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

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

prove

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

True

Solution steps

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

Manipulating right side

1 - csc^{2} (x)

Express with sin, cos

1 - csc^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}

= 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 )}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

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

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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