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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply

cos (θ) \frac{cos ( θ )}{sin ( θ )} : \frac{cos ^{2} ( θ )}{sin ( θ )}

cos (θ) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Apply the fraction rule:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

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

Expand

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

- sin (θ) (1 - sin (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - sin (θ), b = 1, c = sin (θ)

= - sin (θ) \cdot 1 - (- sin (θ)) sin (θ)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot sin (θ) + sin (θ) sin (θ)

Simplify

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

- 1 \cdot sin (θ) + sin (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = 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)

= - sin (θ) + sin^{2} (θ)

= - sin (θ) + sin^{2} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

- sin^{2} (θ) + sin^{2} (θ) = 0

= - sin (θ) + 1

= - sin (θ) + 1

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

Cancel the common factor:

- sin (θ) + 1

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove csc(θ)-sin(θ)=cos(θ)cot(θ)

p ro v e csc (θ) - sin (θ) = cos (θ) cot (θ)

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

p ro v e sec (x) (1 - sin^{2} (x)) = cos (x)

prove (1-cos^2(θ))(1+cot^2(θ))=1

p ro v e (1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1

prove cot(x)sin(x)=cos(x)

p ro v e cot (x) sin (x) = cos (x)

prove sin(2x)+cos(2x)-1=2sin(x)(cos(x)-sin(x))

p ro v e sin (2 x) + cos (2 x) - 1 = 2 sin (x) (cos (x) - sin (x))

Frequently Asked Questions (FAQ)

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