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

The answer to whether (sec(x))/(1-cos(x))=(sec(x)+1)/(sin^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))/(1-cos(x))=(sec(x)+1)/(sin^2(x))

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Multiply by

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Use the Pythagorean identity:

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

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

= \frac{sec ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Apply the fraction rule:

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

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

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

Expand

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

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

Expand

(1 + cos (x)) sec (x) : sec (x) + sec (x) cos (x)

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = sec (x), b = 1, c = cos (x)

= sec (x) \cdot 1 + sec (x) cos (x)

= 1 \cdot sec (x) + sec (x) cos (x)

Multiply:

1 \cdot sec (x) = sec (x)

= sec (x) + sec (x) cos (x)

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

sec (x) cos (x) = 1

sec (x) cos (x)

Express with sin, cos

sec (x) cos (x)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

= 1

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

Apply rule

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(1 - sin (x)) (1 + csc (x)) = csc (x) - sin (x)

1 - cos^{2} (θ) = sin^{2} (θ)

tan (\frac{π}{4} - θ) = \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

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

csc (B) - sin (B) = cot (B) cos (B)

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