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

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

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prove (sec(α))/(1-cos(α))=(sec(α)+1)/(sin^2(α))

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ( α )}{1 - cos ( α )}

Rewrite using trig identities

\frac{sec ( α )}{1 - cos ( α )}

Multiply by

\frac{1 + cos ( α )}{1 + cos ( α )}

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

Apply Difference of Two Squares Formula:

α^{2} - y^{2} = (α + y) (α - y)

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

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

Use the Pythagorean identity:

1 = cos^{2} (α) + sin^{2} (α)

1 - cos^{2} (α) = sin^{2} (α)

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

Apply the fraction rule:

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

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

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

Expand

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

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

Expand

(1 + cos (α)) sec (α) : sec (α) + sec (α) cos (α)

(1 + cos (α)) sec (α)

= sec (α) (1 + cos (α))

Apply the distributive law:

a (b + c) = ab + a c

a = sec (α), b = 1, c = cos (α)

= sec (α) \cdot 1 + sec (α) cos (α)

= 1 \cdot sec (α) + sec (α) cos (α)

Multiply:

1 \cdot sec (α) = sec (α)

= sec (α) + sec (α) cos (α)

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

sec (α) cos (α) = 1

sec (α) cos (α)

Express with sin, cos

sec (α) cos (α)

Use the basic trigonometric identity:

sec (α) = \frac{1}{cos ( α )}

= \frac{1}{cos ( α )} cos (α)

\frac{1}{cos ( α )} cos (α) = 1

\frac{1}{cos ( α )} cos (α)

Multiply fractions:

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

= \frac{1 cos ( α )}{cos ( α )}

Cancel the common factor:

cos (α)

= 1

= 1

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

Apply rule

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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