Is 1/(cos(x)-cot(x))-(sec(x))/(sin(x)-1)=sec(x) ?

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

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prove 1/(cos(x)-cot(x))-(sec(x))/(sin(x)-1)=sec(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Join

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

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

Convert element to fraction:

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

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

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

Apply the fraction rule:

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

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

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

Factor

cos (x) sin (x) - cos (x) : cos (x) (sin (x) - 1)

cos (x) sin (x) - cos (x)

Factor out common term

cos (x)

= cos (x) (sin (x) - 1)

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

Since the denominators are equal, combine the fractions:

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

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

Cancel the common factor:

sin (x) - 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{x}{2}) = \frac{1 + sin ( x )}{2}

\frac{sec ( θ ) + csc ( θ )}{1 + tan ( θ )} = csc (θ)

(cot (a) + tan (a)) cos (a) = csc (a)

(sin^{2} (x) - 1) = - cos^{2} (x)

cos (- \frac{x π}{3}) = cos (\frac{x π}{3})

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