Is tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1) ?

The answer to whether tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1) is True

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prove tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1)

Solution

prove

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

Solution

True

Solution steps

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

Let:

u = \frac{θ}{2}

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

Prove

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

True

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Cancel the common factor:

cos (2 u)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Add the numbers:

1 + 1 = 2

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

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 ^{2} ( u ) + 2}

Factor

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

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

Rewrite as

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

Factor out common term

2

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Remove parentheses:

(- a) = - a

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

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (u) tan (u)

Simplify

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

tan (u) tan (u)

Apply exponent rule:

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

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

= tan^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= tan^{2} (u)

tan^{2} (u)

tan^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1) ?
The answer to whether tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1) is True