Is cot^2(x)-cos^2(x)cot^2(x)=cos^2(x) ?

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

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prove cot^2(x)-cos^2(x)cot^2(x)=cos^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

(\frac{cos ( x )}{sin ( x )})^{2} - cos^{2} (x) (\frac{cos ( x )}{sin ( x )})^{2} : \frac{cos ^{2} ( x ) - cos ^{4} ( x )}{sin ^{2} ( x )}

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

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

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

Apply exponent rule:

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

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

cos^{2} (x) (\frac{cos ( x )}{sin ( x )})^{2} = \frac{cos ^{4} ( x )}{sin ^{2} ( x )}

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

cos^{2} (x) cos^{2} (x) = cos^{4} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= cos^{4} (x)

= \frac{cos ^{4} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )} - \frac{cos ^{4} ( x )}{sin ^{2} ( x )}

Apply rule

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

= \frac{cos ^{2} ( x ) - cos ^{4} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x ) - cos ^{4} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x ) - cos ^{4} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

\frac{cos ^{2} ( x ) - cos ^{4} ( x )}{sin ^{2} ( x )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( x ) - cos ^{4} ( x )}{1 - cos ^{2} ( x )}

Simplify

\frac{cos ^{2} ( x ) - cos ^{4} ( x )}{1 - cos ^{2} ( x )} : cos^{2} (x)

\frac{cos ^{2} ( x ) - cos ^{4} ( x )}{1 - cos ^{2} ( x )}

Factor

cos^{2} (x) - cos^{4} (x) : cos^{2} (x) (1 - cos^{2} (x))

cos^{2} (x) - cos^{4} (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{4} (x) = cos^{2} (x) cos^{2} (x)

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

Factor out common term

cos^{2} (x)

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

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

Cancel the common factor:

1 - cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (\frac{1}{x}) = cos (\frac{1}{x}) arcsin (\frac{1}{x})

\frac{tan ( x )}{sin ( x )} - sec (x) = 0

\frac{sec ^{2} ( y )}{tan ( y )} = tan (y) + cot (y)

cos (2 0^{\circ}) = - sin^{2} (θ)

tan (α) + \frac{cos ( α )}{1 + sin ( α )} = sec (α)

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