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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

\frac{- cot ( x ) + csc ( x )}{csc ( 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 )}{csc ( 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 )}}{\frac{1}{s i n ( x )}}

Simplify

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin (x)

= - - cos (x) + 1

= - cos (x) + 1

= 1 - cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{7 cot ( x )}{sec ( x )} = 7 csc (x) - 7 sin (x)

sin (θ + \frac{π}{3}) = sin (θ) + sin (\frac{π}{3})

(4 sin (θ) - 4 cos (θ))^{2} = 16 - 16 sin (2 θ)

arcsec (x) = cos (x)

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