Is (tan(θ))/(tan(θ)-1)= 1/(1-cot(θ)) ?

The answer to whether (tan(θ))/(tan(θ)-1)= 1/(1-cot(θ)) is True

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Popular Trigonometry >

prove (tan(θ))/(tan(θ)-1)= 1/(1-cot(θ))

Solution

prove

\frac{tan ( θ )}{tan ( θ ) - 1} = \frac{1}{1 - cot ( θ )}

Solution

True

Solution steps

\frac{tan ( θ )}{tan ( θ ) - 1} = \frac{1}{1 - cot ( θ )}

Manipulating left side

\frac{tan ( θ )}{tan ( θ ) - 1}

Express with sin, cos

\frac{tan ( θ )}{- 1 + tan ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

- 1 + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= - - cos (θ) + sin (θ)

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

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

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

Manipulating right side

\frac{1}{1 - cot ( θ )}

Express with sin, cos

\frac{1}{1 - cot ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 - \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Multiply:

1 \cdot sin (θ) = sin (θ)

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

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

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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