Is (cot(θ)-1)/(1-tan(θ))=cot(θ) ?

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

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prove (cot(θ)-1)/(1-tan(θ))=cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{cot ( θ ) - 1}{1 - tan ( θ )}

Express with sin, cos

\frac{- 1 + cot ( θ )}{1 - tan ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Cancel the common factor:

- sin (θ) + cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

6 cos^{2} (B) - 6 sin^{2} (B) = 12 cos^{2} (B) - 6

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

2 sin (x) + csc (x) = 0

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