Is tan^2(x)cot^2(x)=1 ?

The answer to whether tan^2(x)cot^2(x)=1 is True

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prove tan^2(x)cot^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x) cot^{2} (x)

Express with sin, cos

cot^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

Cancel the common factor:

sin^{2} (x)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

cos (π + θ) = - cos (θ)

cos (\frac{3 π}{2} - θ) = - sin (θ)

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

cos (B) csc (B) tan (B) = 1

\frac{tan ( θ ) + cot ( θ )}{tan ( θ )} = csc^{2} (θ)