Is csc^2(x/2)=(2sec(x))/(sec(x)-1) ?

The answer to whether csc^2(x/2)=(2sec(x))/(sec(x)-1) is True

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prove csc^2(x/2)=(2sec(x))/(sec(x)-1)

Solution

prove

csc^{2} (\frac{x}{2}) = \frac{2 sec ( x )}{sec ( x ) - 1}

Solution

True

Solution steps

csc^{2} (\frac{x}{2}) = \frac{2 sec ( x )}{sec ( x ) - 1}

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

Cancel the common factor:

cos (2 u)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Divide the numbers:

\frac{2}{2} = 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{( \frac{1}{c s c ( u )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( u )} ) ^{2}}

(\frac{1}{csc ( u )})^{2} = \frac{1}{csc ^{2} ( u )}

(\frac{1}{csc ( u )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( u )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( u )}

= \frac{1}{\frac{1}{c s c ^{2} ( u )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( u )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (u)

csc^{2} (u)

csc^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

csc^{2} (\frac{x}{2}) = \frac{2 sec ( x )}{sec ( x ) - 1}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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