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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{1 - tan ( θ )}{1 + tan ( θ )}

Express with sin, cos

\frac{1 - tan ( θ )}{1 + tan ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Cancel the common factor:

cos (θ)

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

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

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

Manipulating right side

\frac{cot ( θ ) - 1}{cot ( θ ) + 1}

Express with sin, cos

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

Simplify

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

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

Divide fractions:

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

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

Cancel the common factor:

sin (θ)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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