Is (cos^2(-x)csc(x))/(cot(x))=cos(x) ?

The answer to whether (cos^2(-x)csc(x))/(cot(x))=cos(x) is True

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prove (cos^2(-x)csc(x))/(cot(x))=cos(x)

prove

\frac{cos ^{2} ( - x ) csc ( x )}{cot ( x )} = cos (x)

True

Solution steps

\frac{cos ^{2} ( - x ) csc ( x )}{cot ( x )} = cos (x)

Manipulating left side

\frac{cos ^{2} ( - x ) csc ( x )}{cot ( x )}

Use the negative angle identity:

cos (- x) = cos (x)

= \frac{cos ^{2} ( x ) csc ( x )}{cot ( x )}

Express with sin, cos

\frac{cos ^{2} ( x ) csc ( x )}{cot ( x )}

Use the basic trigonometric identity:

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

= \frac{cos ^{2} ( x ) \frac{1}{s i n ( x )}}{cot ( x )}

Use the basic trigonometric identity:

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

= \frac{cos ^{2} ( x ) \frac{1}{s i n ( x )}}{\frac{c o s ( x )}{s i n ( x )}}

Simplify

\frac{cos ^{2} ( x ) \frac{1}{s i n ( x )}}{\frac{c o s ( x )}{s i n ( x )}} : cos (x)

\frac{cos ^{2} ( x ) \frac{1}{s i n ( x )}}{\frac{c o s ( x )}{s i n ( x )}}

Apply the fraction rule:

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

= \frac{cos ^{2} ( x ) \frac{1}{s i n ( x )} sin ( x )}{cos ( x )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 1 \cdot cos^{2} (x)

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= cos^{2} (x)

= \frac{cos ^{2} ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cot (2 x) = \frac{csc ^{2} ( x ) - 2}{2 cot ( x )}

p ro v e \frac{1}{cos ( θ )} = \frac{tan ( θ )}{sin ( θ )}

p ro v e sin (x) \cdot tan (x) = sec (x) - cos (x)

p ro v e (tan (y) + cot (y)) sin (y) cos (y) = 1

p ro v e (csc (θ) - cot (θ)) (csc (θ) + cot (θ)) = 1

Is (cos^2(-x)csc(x))/(cot(x))=cos(x) ?
The answer to whether (cos^2(-x)csc(x))/(cot(x))=cos(x) is True