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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

= \frac{- cos ( θ ) + 1}{sin ( θ )}

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

Apply the fraction rule:

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

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

Join

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

- 1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{- cos ( θ ) + 1}{cos ( θ )}

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

- cos (θ) + 1

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (z) + cos^{2} (z) = 1

tan (x) sin (x) = cos (x)

cot (θ) = \frac{1}{tan ( θ )}

cos^{2} (θ) tan^{2} (θ) = sin^{2} (θ)

1 - cot (x) = csc^{2} (x) - 2 cot (x)

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