Is 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x) ?

The answer to whether 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x) is True

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Popular Trigonometry >

prove 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x)

Solution

prove

\frac{1}{1 - cos ( x )} - \frac{1}{1 + cos ( x )} = 2 cot (x) csc (x)

Solution

True

Solution steps

\frac{1}{1 - cos ( x )} - \frac{1}{1 + cos ( x )} = 2 cot (x) csc (x)

Manipulating left side

\frac{1}{1 - cos ( x )} - \frac{1}{1 + cos ( x )}

Simplify

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

- \frac{1}{1 + cos ( x )} + \frac{1}{1 - cos ( x )}

Least Common Multiplier of

1 + cos (x), 1 - cos (x) : (cos (x) + 1) (- cos (x) + 1)

1 + cos (x), 1 - cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + cos (x)

1 - cos (x)

= (cos (x) + 1) (- cos (x) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

(cos (x) + 1) (- cos (x) + 1)

For

\frac{1}{1 + cos ( x )} :

multiply the denominator and numerator by

- cos (x) + 1

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

For

\frac{1}{1 - cos ( x )} :

multiply the denominator and numerator by

cos (x) + 1

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

= - \frac{- cos ( x ) + 1}{( cos ( x ) + 1 ) ( - cos ( x ) + 1 )} + \frac{cos ( x ) + 1}{( cos ( x ) + 1 ) ( - cos ( x ) + 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{- ( - cos ( x ) + 1 ) + cos ( x ) + 1}{( cos ( x ) + 1 ) ( - cos ( x ) + 1 )}

Expand

- (- cos (x) + 1) + cos (x) + 1 : 2 cos (x)

- (- cos (x) + 1) + cos (x) + 1

- (- cos (x) + 1) : cos (x) - 1

- (- cos (x) + 1)

Distribute parentheses

= - (- cos (x)) - (1)

Apply minus-plus rules

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

= cos (x) - 1

= cos (x) - 1 + cos (x) + 1

Simplify

cos (x) - 1 + cos (x) + 1 : 2 cos (x)

cos (x) - 1 + cos (x) + 1

Group like terms

= cos (x) + cos (x) - 1 + 1

Add similar elements:

cos (x) + cos (x) = 2 cos (x)

= 2 cos (x) - 1 + 1

- 1 + 1 = 0

= 2 cos (x)

= 2 cos (x)

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

2 cot (x) csc (x)

Express with sin, cos

2 cot (x) csc (x)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{cos ( x )}{sin ( x )} csc (x)

Use the basic trigonometric identity:

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

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

Simplify

2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )} : \frac{2 cos ( x )}{sin ^{2} ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x) ?
The answer to whether 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x) is True