Is (sec(x)-1)/(sin(x)sec(x))=tan(x/2) ?

The answer to whether (sec(x)-1)/(sin(x)sec(x))=tan(x/2) is True

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prove (sec(x)-1)/(sin(x)sec(x))=tan(x/2)

Solution

prove

\frac{sec ( x ) - 1}{sin ( x ) sec ( x )} = tan (\frac{x}{2})

Solution

True

Solution steps

\frac{sec ( x ) - 1}{sin ( x ) sec ( x )} = tan (\frac{x}{2})

Let:

u = \frac{x}{2}

\frac{sec ( 2 u ) - 1}{sin ( 2 u ) sec ( 2 u )} = tan (u)

Prove

\frac{sec ( 2 u ) - 1}{sin ( 2 u ) sec ( 2 u )} = tan (u) :

True

\frac{sec ( 2 u ) - 1}{sin ( 2 u ) sec ( 2 u )} = tan (u)

Manipulating left side

\frac{sec ( 2 u ) - 1}{sin ( 2 u ) sec ( 2 u )}

Express with sin, cos

\frac{- 1 + sec ( 2 u )}{sec ( 2 u ) sin ( 2 u )}

Use the basic trigonometric identity:

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

= \frac{- 1 + \frac{1}{c o s ( 2 u )}}{\frac{1}{c o s ( 2 u )} sin ( 2 u )}

Simplify

\frac{- 1 + \frac{1}{c o s ( 2 u )}}{\frac{1}{c o s ( 2 u )} sin ( 2 u )} : \frac{- cos ( 2 u ) + 1}{sin ( 2 u )}

\frac{- 1 + \frac{1}{c o s ( 2 u )}}{\frac{1}{c o s ( 2 u )} sin ( 2 u )}

Multiply

\frac{1}{cos ( 2 u )} sin (2 u) : \frac{sin ( 2 u )}{cos ( 2 u )}

\frac{1}{cos ( 2 u )} sin (2 u)

Multiply fractions:

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

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

Multiply:

1 \cdot sin (2 u) = sin (2 u)

= \frac{sin ( 2 u )}{cos ( 2 u )}

= \frac{- 1 + \frac{1}{c o s ( 2 u )}}{\frac{s i n ( 2 u )}{c o s ( 2 u )}}

Join

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

- 1 + \frac{1}{cos ( 2 u )}

Convert element to fraction:

1 = \frac{1 cos ( 2 u )}{cos ( 2 u )}

= - \frac{1 \cdot cos ( 2 u )}{cos ( 2 u )} + \frac{1}{cos ( 2 u )}

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 u ) + 1}{cos ( 2 u )}

Multiply:

1 \cdot cos (2 u) = cos (2 u)

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

= \frac{\frac{- c o s ( 2 u ) + 1}{c o s ( 2 u )}}{\frac{s i n ( 2 u )}{c o s ( 2 u )}}

Divide fractions:

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

= \frac{( - cos ( 2 u ) + 1 ) cos ( 2 u )}{cos ( 2 u ) sin ( 2 u )}

Cancel the common factor:

cos (2 u)

= \frac{- cos ( 2 u ) + 1}{sin ( 2 u )}

= \frac{1 - cos ( 2 u )}{sin ( 2 u )}

= \frac{1 - cos ( 2 u )}{sin ( 2 u )}

Rewrite using trig identities

\frac{1 - cos ( 2 u )}{sin ( 2 u )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{1 - cos ( 2 u )}{2 sin ( u ) cos ( u )}

Use the Double Angle identity:

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

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

Simplify

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

\frac{1 - ( 1 - 2 sin ^{2} ( u ) )}{2 sin ( u ) cos ( u )}

Expand

1 - (1 - 2 sin^{2} (u)) : 2 sin^{2} (u)

1 - (1 - 2 sin^{2} (u))

- (1 - 2 sin^{2} (u)) : - 1 + 2 sin^{2} (u)

- (1 - 2 sin^{2} (u))

Distribute parentheses

= - (1) - (- 2 sin^{2} (u))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (u)

= 1 - 1 + 2 sin^{2} (u)

1 - 1 = 0

= 2 sin^{2} (u)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (u)

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

Use the basic trigonometric identity:

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

= tan (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

\frac{sec ( x ) - 1}{sin ( x ) sec ( x )} = tan (\frac{x}{2})

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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