Is (cot(-θ))/(csc(θ))=-cos(θ) ?

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

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prove (cot(-θ))/(csc(θ))=-cos(θ)

prove

\frac{cot ( - θ )}{csc ( θ )} = - cos (θ)

True

Solution steps

\frac{cot ( - θ )}{csc ( θ )} = - cos (θ)

Manipulating left side

\frac{cot ( - θ )}{csc ( θ )}

Use the negative angle identity:

cot (- x) = - cot (x)

= \frac{- cot ( θ )}{csc ( θ )}

Express with sin, cos

\frac{- cot ( θ )}{csc ( 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 )}

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 )}}

Simplify

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

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin (θ)

= - cos (θ)

= - cos (θ)

= - cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

sec (α) - cos (α) = sin (α) tan (α)

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

tan^{2} (θ) + 6 = sec^{2} (θ) + 5

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

sec (\frac{π}{2} - x) = csc (x)

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