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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

tan (x) (1 - cot (x))

Express with sin, cos

(1 - cot (x)) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

(1 - \frac{cos ( x )}{sin ( x )}) \frac{sin ( x )}{cos ( x )} : \frac{sin ( x ) - cos ( x )}{cos ( x )}

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

Multiply fractions:

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= sin (x) - cos (x)

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

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

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

Manipulating right side

tan (x) - 1

Express with sin, cos

- 1 + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

- 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{- cos ( x ) + sin ( x )}{cos ( x )}

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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