Is (1-cos(x))(1+sec(x))=sec(x)-cos(x) ?

The answer to whether (1-cos(x))(1+sec(x))=sec(x)-cos(x) is True

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prove (1-cos(x))(1+sec(x))=sec(x)-cos(x)

Solution

prove

(1 - cos (x)) (1 + sec (x)) = sec (x) - cos (x)

Solution

True

Solution steps

(1 - cos (x)) (1 + sec (x)) = sec (x) - cos (x)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Multiply fractions:

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

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

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

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

Manipulating right side

sec (x) - cos (x)

Express with sin, cos

- cos (x) + sec (x)

Use the basic trigonometric identity:

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

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

Simplify

- 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{- cos ^{2} ( x ) + 1}{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 (1-cos(x))(1+sec(x))=sec(x)-cos(x) ?
The answer to whether (1-cos(x))(1+sec(x))=sec(x)-cos(x) is True