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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

sin (x) - cos (x) = - (cos (x) - sin (x))

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

Refine

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

Cancel the common factor:

cos (x) - sin (x)

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

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

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

Use the basic trigonometric identity:

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

= - tan (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

p ro v e 1 + cot^{2} (- x) = csc^{2} (x)

p ro v e cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

p ro v e 5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

p ro v e csc (2 x) = \frac{sec ( x ) csc ( x )}{2}

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