Is cot^2(v/2)=((sec(v)+1))/((sec(v)-1)) ?

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

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Popular Trigonometry >

prove cot^2(v/2)=((sec(v)+1))/((sec(v)-1))

Solution

prove

cot^{2} (\frac{v}{2}) = \frac{( sec ( v ) + 1 )}{( sec ( v ) - 1 )}

Solution

True

Solution steps

cot^{2} (\frac{v}{2}) = \frac{sec ( v ) + 1}{sec ( v ) - 1}

Let:

u = \frac{v}{2}

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

Prove

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

True

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

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

Simplify

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

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

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

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (u)

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Add the numbers:

1 + 1 = 2

= - 2 cos^{2} (u) + 2

= - 2 cos^{2} (u) + 2

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

Factor

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

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

Rewrite as

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

Factor out common term

2

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Remove parentheses:

(- a) = - a

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

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

cot (u) cot (u)

Simplify

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

cot (u) cot (u)

Apply exponent rule:

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

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

= cot^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= cot^{2} (u)

cot^{2} (u)

cot^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

cot^{2} (\frac{v}{2}) = \frac{sec ( v ) + 1}{sec ( v ) - 1}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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