Is prove-csc(x)cos(x)=-cot(x) ?

The answer to whether prove-csc(x)cos(x)=-cot(x) is True

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prove-csc(x)cos(x)=-cot(x)

prove

- csc (x) cos (x) = - cot (x)

True

Solution steps

- csc (x) cos (x) = - cot (x)

Manipulating left side

- csc (x) cos (x)

Rewrite using trig identities

- csc (x) cos (x)

Express with sin, cos

csc (x) cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

= \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)

= - cot (x)

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

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

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

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