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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (sec(x)-cos(x))/(cos(x))=tan^2(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (x) tan (x)

Simplify

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

Apply exponent rule:

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

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

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 1-cos^4(x)=2sin^2(x)-sin^4(x)prove

1 - cos^{4} (x) = 2 sin^{2} (x) - sin^{4} (x)

prove tan(pi/2-x)sin(x)=cos(x)prove

tan (\frac{π}{2} - x) sin (x) = cos (x)

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

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

prove 1-tan^2(x/2)=(2cos(x))/(1+cos(x))prove

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

prove (1+cos(θ))/(1+sec(θ))=cos(θ)prove

\frac{1 + cos ( θ )}{1 + sec ( θ )} = cos (θ)

Frequently Asked Questions (FAQ)

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