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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

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

2 sin (2 x) sin (2 x - \frac{π}{2}) = - cos (4 x - \frac{π}{2})

cot (4 5^{\circ}) = \frac{cos ( 4 5 ^{\circ} )}{sin ( 4 5 ^{\circ} )}

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

\frac{1 - tan ^{2} ( x )}{1 + tan ^{2} ( x )} = tan (x)

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