Is (sec^2(x)-1)cot^2(x)=1 ?

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

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prove (sec^2(x)-1)cot^2(x)=1

prove

(sec^{2} (x) - 1) cot^{2} (x) = 1

True

Solution steps

(sec^{2} (x) - 1) cot^{2} (x) = 1

Manipulating left side

(sec^{2} (x) - 1) cot^{2} (x)

Express with sin, cos

(- 1 + sec^{2} (x)) cot^{2} (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply fractions:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

\frac{1 + cos ( θ )}{tan ( θ ) + sin ( θ )} = cot (θ)

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

\frac{4 cot ( x )}{sec ( x )} = 4 csc (x) - 4 sin (x)

(tan (x) sec (- x)) cos (x) = sin (x) - 1

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