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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(- a) = - a

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Cancel the common factor:

sin^{2} (x)

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

= \frac{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 csc^2(x)(sec^2(x)-1)= 1/(cos^2(x)) ?
The answer to whether csc^2(x)(sec^2(x)-1)= 1/(cos^2(x)) is True