Is 1/(2cot(x)sin^2(x))=csc(2x) ?

The answer to whether 1/(2cot(x)sin^2(x))=csc(2x) is True

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prove 1/(2cot(x)sin^2(x))=csc(2x)

prove

\frac{1}{2 cot ( x ) sin ^{2} ( x )} = csc (2 x)

True

Solution steps

\frac{1}{2 cot ( x ) sin ^{2} ( x )} = csc (2 x)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 2 cos (x) sin (x)

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

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

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

Manipulating right side

csc (2 x)

Express with sin, cos

csc (2 x)

Use the basic trigonometric identity:

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ^{2} ( x )}{sec ( x ) - 1} = 1 + \frac{1}{cos ( x )}

cos (α + β) - cos (α - β) = - 2 sin (α) cos (β)

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

7 sec (θ) = \frac{7 sec ( θ ) - 7}{1 - cos ( θ )}

2 tan (x) sin (x) - 3 cos (x) = 2

Is 1/(2cot(x)sin^2(x))=csc(2x) ?
The answer to whether 1/(2cot(x)sin^2(x))=csc(2x) is True