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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

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

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

tan (x) (cot (x) + tan (x)) = sec^{2} (x)

\frac{1 + sin ( 2 x )}{sin ( 2 x )} = 1 + \frac{1}{2} sec (x) csc (x)

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

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

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