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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1}{( - cos ( x ) + \frac{1}{c o s ( x )} ) \frac{1}{c o s ( x )}}

Multiply

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(- a) = - a

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

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

Join

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

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

Convert element to fraction:

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

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

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

- cos (x) cos (x) + 1

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Apply the fraction rule:

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

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

Manipulating right side

cot^{2} (x)

Express with sin, cos

cot^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

Rewrite using trig identities

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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