Is 2csc^2(y)= 1/(1-cos(y))+1/(1+cos(y)) ?

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

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Popular Trigonometry >

prove 2csc^2(y)= 1/(1-cos(y))+1/(1+cos(y))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Simplify

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

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

Least Common Multiplier of

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

1 + cos (y), 1 - cos (y)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + cos (y)

1 - cos (y)

= (cos (y) + 1) (- cos (y) + 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 (y) + 1) (- cos (y) + 1)

For

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

multiply the denominator and numerator by

- cos (y) + 1

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

For

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

multiply the denominator and numerator by

cos (y) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

- cos (y) + 1 + cos (y) + 1 = 2

- cos (y) + 1 + cos (y) + 1

Group like terms

= - cos (y) + cos (y) + 1 + 1

Add similar elements:

- cos (y) + cos (y) = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (y)) (1 - cos (y))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (y)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (y)

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

Use the Pythagorean identity:

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

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

= \frac{2}{sin ^{2} ( y )}

= \frac{2}{sin ^{2} ( y )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= 2 csc^{2} (y)

2 csc^{2} (y)

2 csc^{2} (y)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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