Is (cot(θ)-csc(θ))/(cot(θ)+csc(θ))=(1-2cos(θ)+cos^2(θ))/(-sin^2(θ)) ?

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

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prove (cot(θ)-csc(θ))/(cot(θ)+csc(θ))=(1-2cos(θ)+cos^2(θ))/(-sin^2(θ))

Solution

prove

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )} = \frac{1 - 2 cos ( θ ) + cos ^{2} ( θ )}{- sin ^{2} ( θ )}

Solution

True

Solution steps

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )} = \frac{1 - 2 cos ( θ ) + cos ^{2} ( θ )}{- sin ^{2} ( θ )}

Manipulating left side

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )}

Express with sin, cos

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - csc ( θ )}{\frac{c o s ( θ )}{s i n ( θ )} + csc ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}}

Simplify

\frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}} : \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

\frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}}

Combine the fractions

\frac{cos ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )} : \frac{cos ( θ ) + 1}{sin ( θ )}

Apply rule

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

= \frac{cos ( θ ) + 1}{sin ( θ )}

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ ) + 1}{s i n ( θ )}}

Combine the fractions

\frac{cos ( θ )}{sin ( θ )} - \frac{1}{sin ( θ )} : \frac{cos ( θ ) - 1}{sin ( θ )}

Apply rule

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

= \frac{cos ( θ ) - 1}{sin ( θ )}

= \frac{\frac{c o s ( θ ) - 1}{s i n ( θ )}}{\frac{c o s ( θ ) + 1}{s i n ( θ )}}

Divide fractions:

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

= \frac{( cos ( θ ) - 1 ) sin ( θ )}{sin ( θ ) ( cos ( θ ) + 1 )}

Cancel the common factor:

sin (θ)

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Remove parentheses:

(a) = a

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

Cancel

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

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

Factor

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

1 - 2 cos (θ) + cos^{2} (θ)

Write in the standard form

a x^{2} + b x + c

= cos^{2} (θ) - 2 cos (θ) + 1

Rewrite

cos^{2} (θ) - 2 cos (θ) + 1

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

cos^{2} (θ) - 2 cos (θ) + 1

Rewrite

1

1^{2}

= cos^{2} (θ) - 2 cos (θ) + 1^{2}

Rewrite

2 cos (θ)

2 cos (θ) \cdot 1

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

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

Apply Perfect Square Formula:

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

a = cos (θ), b = 1

= (cos (θ) - 1)^{2}

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

Factor

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

1 - cos^{2} (θ)

Factor out common term

- 1

= - (cos^{2} (θ) - 1)

Factor

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

cos^{2} (θ) - 1

Rewrite

1

1^{2}

= cos^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

Cancel the common factor:

cos (θ) - 1

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

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

Apply rule

- (- a) = a

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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