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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

- \frac{cos ( a )}{sin ( a )} + \frac{sin ( a )}{1 - cos ( a )}

Least Common Multiplier of

sin (a), 1 - cos (a) : sin (a) (- cos (a) + 1)

sin (a), 1 - cos (a)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (a)

1 - cos (a)

= sin (a) (- cos (a) + 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

sin (a) (- cos (a) + 1)

For

\frac{cos ( a )}{sin ( a )} :

multiply the denominator and numerator by

- cos (a) + 1

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

For

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

multiply the denominator and numerator by

sin (a)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

1 - cos^{2} (a) - (1 - cos (a)) cos (a)

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

Expand

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

- cos (a) (1 - cos (a))

Apply the distributive law:

a (b - c) = ab - a c

a = - cos (a), b = 1, c = cos (a)

= - cos (a) \cdot 1 - (- cos (a)) cos (a)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot cos (a) + cos (a) cos (a)

Simplify

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

- 1 \cdot cos (a) + cos (a) cos (a)

1 \cdot cos (a) = cos (a)

1 \cdot cos (a)

Multiply:

1 \cdot cos (a) = cos (a)

= cos (a)

cos (a) cos (a) = cos^{2} (a)

cos (a) cos (a)

Apply exponent rule:

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

cos (a) cos (a) = cos^{1 + 1} (a)

= cos^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cos^{2} (a)

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

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

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

Simplify

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

1 - cos^{2} (a) - cos (a) + cos^{2} (a)

Group like terms

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

Add similar elements:

- cos^{2} (a) + cos^{2} (a) = 0

= - cos (a) + 1

= - cos (a) + 1

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

Cancel the common factor:

- cos (a) + 1

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( a )}}

Simplify

\frac{1}{\frac{1}{c s c ( a )}}

Apply the fraction rule:

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

= \frac{csc ( a )}{1}

Apply rule

\frac{a}{1} = a

= csc (a)

csc (a)

csc (a)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 2sin^2(a)+cos^2(a)=1+sin^2(a)prove

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

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

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

prove sin(2t)=2sin(t)cos(t)prove

sin (2 t) = 2 sin (t) cos (t)

prove sin(2a)+cos(a)=cos(a)(2sin(a)+1)prove

sin (2 a) + cos (a) = cos (a) (2 sin (a) + 1)

prove (csc(x))/(2cos(x))=csc(2x)prove

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

Frequently Asked Questions (FAQ)

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