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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

sin (x) + cos (x) = \frac{cot ( x ) + 1}{csc ( x )}

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

sin (\frac{2 π}{3}) = sin (\frac{- 11 π}{3})

cos^{2} (5 θ) - sin^{2} (5 θ) = cos (10 θ)

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

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