Is (csc(t)+1)/(cot(t))=(cot(t))/(csc(t)-1) ?

The answer to whether (csc(t)+1)/(cot(t))=(cot(t))/(csc(t)-1) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (csc(t)+1)/(cot(t))=(cot(t))/(csc(t)-1)

Solution

prove

\frac{csc ( t ) + 1}{cot ( t )} = \frac{cot ( t )}{csc ( t ) - 1}

Solution

True

Solution steps

\frac{csc ( t ) + 1}{cot ( t )} = \frac{cot ( t )}{csc ( t ) - 1}

Manipulating left side

\frac{csc ( t ) + 1}{cot ( t )}

Express with sin, cos

\frac{1 + csc ( t )}{cot ( t )}

Use the basic trigonometric identity:

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

= \frac{1 + \frac{1}{s i n ( t )}}{cot ( t )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

= \frac{( 1 + \frac{1}{s i n ( t )} ) sin ( t )}{cos ( t )}

Join

1 + \frac{1}{sin ( t )} : \frac{sin ( t ) + 1}{sin ( t )}

1 + \frac{1}{sin ( t )}

Convert element to fraction:

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

= \frac{1 \cdot sin ( t )}{sin ( t )} + \frac{1}{sin ( t )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ( t ) + 1}{sin ( t )}

Multiply:

1 \cdot sin (t) = sin (t)

= \frac{sin ( t ) + 1}{sin ( t )}

= \frac{\frac{s i n ( t ) + 1}{s i n ( t )} sin ( t )}{cos ( t )}

Multiply

\frac{sin ( t ) + 1}{sin ( t )} sin (t) : sin (t) + 1

\frac{sin ( t ) + 1}{sin ( t )} sin (t)

Multiply fractions:

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

= \frac{( sin ( t ) + 1 ) sin ( t )}{sin ( t )}

Cancel the common factor:

sin (t)

= sin (t) + 1

= \frac{sin ( t ) + 1}{cos ( t )}

= \frac{sin ( t ) + 1}{cos ( t )}

= \frac{1 + sin ( t )}{cos ( t )}

Rewrite using trig identities

Multiply by

\frac{1 - sin ( t )}{1 - sin ( t )}

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

Expand

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

(1 + sin (t)) (1 - sin (t))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (t)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (t)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (t)

= \frac{cos ( t )}{1 - sin ( t )}

= \frac{cos ( t )}{1 - sin ( t )}

Manipulating right side

\frac{cot ( t )}{csc ( t ) - 1}

Express with sin, cos

\frac{cot ( t )}{- 1 + csc ( t )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

- 1 + \frac{1}{sin ( t )} : \frac{- sin ( t ) + 1}{sin ( t )}

- 1 + \frac{1}{sin ( t )}

Convert element to fraction:

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

= - \frac{1 \cdot sin ( t )}{sin ( t )} + \frac{1}{sin ( t )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ( t ) + 1}{sin ( t )}

Multiply:

1 \cdot sin (t) = sin (t)

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

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

Multiply

sin (t) \frac{- sin ( t ) + 1}{sin ( t )} : - sin (t) + 1

sin (t) \frac{- sin ( t ) + 1}{sin ( t )}

Multiply fractions:

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

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

Cancel the common factor:

sin (t)

= - - sin (t) + 1

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

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

= \frac{cos ( t )}{1 - sin ( t )}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (3sin(θ)-3cos(θ))^2=9-9sin(2θ)prove

(3 sin (θ) - 3 cos (θ))^{2} = 9 - 9 sin (2 θ)

prove (cos(x))/(sin^2(x))=csc(x)cot(x)prove

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

prove tan(2x)-tan(x)=tan(x)sec(2x)prove

tan (2 x) - tan (x) = tan (x) sec (2 x)

prove tan(x)sin(x)=sec(x)-cos(x)prove

tan (x) sin (x) = sec (x) - cos (x)

prove (cos(x))^2=cos^2(x)prove

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

Frequently Asked Questions (FAQ)

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