Is (8csc(-x))/(sec(-x))=-8cot(x) ?

The answer to whether (8csc(-x))/(sec(-x))=-8cot(x) is True

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prove (8csc(-x))/(sec(-x))=-8cot(x)

prove

\frac{8 csc ( - x )}{sec ( - x )} = - 8 cot (x)

True

Solution steps

\frac{8 csc ( - x )}{sec ( - x )} = - 8 cot (x)

Manipulating left side

\frac{8 csc ( - x )}{sec ( - x )}

Use the negative angle identity:

csc (- x) = - csc (x)

= \frac{8 ( - csc ( x ) )}{sec ( - x )}

Use the negative angle identity:

sec (- x) = sec (x)

= \frac{8 ( - csc ( x ) )}{sec ( x )}

Express with sin, cos

\frac{( - csc ( x ) ) \cdot 8}{sec ( x )}

Use the basic trigonometric identity:

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

= \frac{( - \frac{1}{s i n ( x )} ) \cdot 8}{sec ( x )}

Use the basic trigonometric identity:

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

= \frac{( - \frac{1}{s i n ( x )} ) \cdot 8}{\frac{1}{c o s ( x )}}

Simplify

\frac{( - \frac{1}{s i n ( x )} ) \cdot 8}{\frac{1}{c o s ( x )}} : - \frac{8 cos ( x )}{sin ( x )}

\frac{( - \frac{1}{s i n ( x )} ) \cdot 8}{\frac{1}{c o s ( x )}}

Remove parentheses:

(- a) = - a

= \frac{- \frac{1}{s i n ( x )} \cdot 8}{\frac{1}{c o s ( x )}}

Apply the fraction rule:

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

= - \frac{\frac{1}{s i n ( x )} \cdot 8}{\frac{1}{c o s ( x )}}

Apply the fraction rule:

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

\frac{\frac{1}{s i n ( x )} \cdot 8}{\frac{1}{c o s ( x )}} = \frac{\frac{1}{s i n ( x )} \cdot 8 cos ( x )}{1}

= - \frac{\frac{1}{s i n ( x )} \cdot 8 cos ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{1}{s i n ( x )} \cdot 8 cos ( x )}{1} = \frac{1}{sin ( x )} \cdot 8 cos (x)

= - \frac{1}{sin ( x )} \cdot 8 cos (x)

Multiply fractions:

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

= - \frac{1 \cdot 8 cos ( x )}{sin ( x )}

Multiply the numbers:

1 \cdot 8 = 8

= - \frac{8 cos ( x )}{sin ( x )}

= - \frac{8 cos ( x )}{sin ( x )}

= - \frac{8 cos ( x )}{sin ( x )}

Rewrite using trig identities

= - 8 \cdot \frac{cos ( x )}{sin ( x )}

Use the basic trigonometric identity:

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

- 8 cot (x)

- 8 cot (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cos (α - β) = cos (α) cos (β) + sin (α) sin (β)

p ro v e cos^{2} (x) - sin^{2} (x) = 1 - sin^{2} (x)

p ro v e sin^{3} (x) = \frac{3}{4} sin (x) - \frac{1}{4} sin (3 x)

p ro v e cos^{2} (x) = (1 - sin (x)) (1 + sin (x))

p ro v e \frac{( cos ( x ) sec ( x ) )}{tan ( x )} = cot (x)

Is (8csc(-x))/(sec(-x))=-8cot(x) ?
The answer to whether (8csc(-x))/(sec(-x))=-8cot(x) is True