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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

\frac{sin ( θ ) - csc ( θ )}{cos ( θ ) - cot ( θ )}

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- \frac{1}{s i n ( θ )} + sin ( θ )}{cos ( θ ) - cot ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

cos (θ) - \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

cos (θ) = \frac{cos ( θ ) sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

- \frac{1}{sin ( θ )} + sin (θ)

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

- 1 + sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= - 1 + sin^{2} (θ)

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

= \frac{\frac{- 1 + s i n ^{2} ( θ )}{s i n ( θ )}}{\frac{c o s ( θ ) s i n ( θ ) - c o s ( θ )}{s i n ( θ )}}

Divide fractions:

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

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

Cancel the common factor:

sin (θ)

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

Factor

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

- 1 + sin^{2} (θ)

Rewrite

1

1^{2}

= sin^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sin (θ) + 1) (sin (θ) - 1)

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

Factor out common term

cos (θ)

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

Cancel the common factor:

sin (θ) - 1

= \frac{sin ( θ ) + 1}{cos ( θ )}

= \frac{sin ( θ ) + 1}{cos ( θ )}

= \frac{1 + sin ( θ )}{cos ( θ )}

Rewrite using trig identities

Multiply by

\frac{1 - sin ( θ )}{1 - sin ( θ )}

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

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

Cancel the common factor:

cos (θ)

= \frac{cos ( θ )}{1 - sin ( θ )}

= \frac{cos ( θ )}{1 - sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove csc^2(x)= 1/(sin^2(x))prove

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

prove (sec^2(θ)-1)cos^2(θ)=sin^2(θ)prove

(sec^{2} (θ) - 1) cos^{2} (θ) = sin^{2} (θ)

prove (sin(θ)+tan(θ))/(1+cos(θ))=tan(θ)prove

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

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

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

Frequently Asked Questions (FAQ)

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