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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (\frac{π}{2} - u) = cot (u)

p ro v e csc (θ) - cot (θ) = \frac{sin ( θ )}{1 + cos ( θ )}

p ro v e tan (2 π - θ) = - tan (θ)

p ro v e sin^{4} (θ) - cos^{4} (θ) = 1 - 2 cos^{2} (θ)

p ro v e \frac{1 + tan ( x )}{1 - tan ( x )} + \frac{1 + cot ( x )}{1 - cot ( x )} = 0

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