Is (csc(x)-cot(x))/(sec(x)-1)=cot(x) ?

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

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prove (csc(x)-cot(x))/(sec(x)-1)=cot(x)

prove

\frac{csc ( x ) - cot ( x )}{sec ( x ) - 1} = cot (x)

True

Solution steps

\frac{csc ( x ) - cot ( x )}{sec ( x ) - 1} = cot (x)

Manipulating left side

\frac{csc ( x ) - cot ( x )}{sec ( x ) - 1}

Express with sin, cos

\frac{- cot ( x ) + csc ( x )}{- 1 + sec ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

- cos (x) + 1

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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