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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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)

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

- cos (θ) + 1 = - (cos (θ) - 1)

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

Refine

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

Cancel the common factor:

cos (θ) - 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove sec(θ)-tan(θ)sin(θ)= 1/(sec(θ))prove

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

prove cot(-a)cos(-a)+sin(-a)=-csc(a)prove

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

prove sin(2a)-tan(a)=tan(a)cos(2a)prove

sin (2 a) - tan (a) = tan (a) cos (2 a)

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

2 cos (x) tan (x) csc (x) = 2

prove sec^2(y)-cot^2(pi/2-y)=1 prove

sec^{2} (y) - cot^{2} (\frac{π}{2} - y) = 1

Frequently Asked Questions (FAQ)

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