Is (-cot(x)+1)/(tan(x)-1)=cot(x) ?

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

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prove (-cot(x)+1)/(tan(x)-1)=cot(x)

prove

\frac{- cot ( x ) + 1}{tan ( x ) - 1} = cot (x)

True

Solution steps

\frac{- cot ( x ) + 1}{tan ( x ) - 1} = cot (x)

Manipulating left side

\frac{- cot ( x ) + 1}{tan ( x ) - 1}

Express with sin, cos

\frac{1 - cot ( x )}{- 1 + tan ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Join

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

1 - \frac{cos ( x )}{sin ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Divide fractions:

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

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

Cancel the common factor:

sin (x) - cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 4 cos^{2} (a) + 4 sin^{2} (a) = 4

p ro v e \frac{sin ( a )}{tan ( a )} = cos (a)

p ro v e sin^{2} (3 x) + cos^{2} (3 x) = 1

p ro v e cos (\frac{π}{7}) = sin (\frac{π}{2} - \frac{π}{7})

p ro v e arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

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