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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Join

1 + \frac{1}{cos ( x )} : \frac{cos ( x ) + 1}{cos ( x )}

1 + \frac{1}{cos ( x )}

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

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

Multiply fractions:

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

= \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

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Frequently Asked Questions (FAQ)

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