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

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

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Popular Trigonometry >

prove (sec^2(x)-1)cos(x)=sin^2(x)sec(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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