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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

csc (- x) = - csc (x)

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

Use the negative angle identity:

sec (- x) = sec (x)

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Divide fractions:

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

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

Refine

= - \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

p ro v e cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

p ro v e sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos (x)

p ro v e (cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 csc^{2} (θ) - 1

p ro v e \frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

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