Is 1/2 (tan(b))/(1-cos^2(b))=csc(2b) ?

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

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prove 1/2 (tan(b))/(1-cos^2(b))=csc(2b)

prove

\frac{1}{2} \frac{tan ( b )}{1 - cos ^{2} ( b )} = csc (2 b)

True

Solution steps

\frac{1}{2} \cdot \frac{tan ( b )}{1 - cos ^{2} ( b )} = csc (2 b)

Manipulating left side

\frac{1}{2} \frac{tan ( b )}{1 - cos ^{2} ( b )}

Express with sin, cos

\frac{1}{2} \cdot \frac{tan ( b )}{1 - cos ^{2} ( b )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot \frac{sin ( b )}{cos ( b )} = \frac{sin ( b )}{cos ( b )}

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (b)

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

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (2 b)

csc (2 b)

csc (2 b)

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{4 + 2 sin ( θ )}{1 - sin ( θ )} = 3

\frac{sin ( 4 x )}{4} = 4 \frac{sin ( x ) cos ( x )}{4}

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

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