Is 1/(cot(x)(1-cos(2x)))=csc(2x) ?

The answer to whether 1/(cot(x)(1-cos(2x)))=csc(2x) is True

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prove 1/(cot(x)(1-cos(2x)))=csc(2x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{cot ( x ) ( 1 - cos ( 2 x ) )}

Rewrite using trig identities

\frac{1}{cot ( x ) ( 1 - cos ( 2 x ) )}

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

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

Expand

1 - (1 - 2 sin^{2} (x)) : 2 sin^{2} (x)

1 - (1 - 2 sin^{2} (x))

- (1 - 2 sin^{2} (x)) : - 1 + 2 sin^{2} (x)

- (1 - 2 sin^{2} (x))

Distribute parentheses

= - (1) - (- 2 sin^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (x)

= 1 - 1 + 2 sin^{2} (x)

1 - 1 = 0

= 2 sin^{2} (x)

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( 2 x )}}

Simplify

\frac{1}{\frac{1}{c s c ( 2 x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc (2 x)

csc (2 x)

csc (2 x)

We showed that the two sides could take the same form

\Rightarrow True

tan^{4} (x) = tan^{2} (x) sec^{2} (x) - tan^{2} (x)

(tan (θ) + cot (θ)) cos (θ) = csc (θ)

cos^{4} (t) - sin^{4} (t) = 1 - 2 sin^{2} (t)

\frac{sec ( x ) + csc ( x )}{1 + tan ( x )} = csc (x)

1 + tan^{2} (- θ) = sec^{2} (θ)

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