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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

Multiply

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

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

Cancel the common factor:

cos (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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