Is csc(2x)= 1/2 sec(x)csc(x) ?

The answer to whether csc(2x)= 1/2 sec(x)csc(x) is True

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prove csc(2x)= 1/2 sec(x)csc(x)

prove

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

True

Solution steps

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

Manipulating left 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 )}

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 \cdot 1 = 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

cos (π - θ) + sin (\frac{π}{2} + θ) = 0

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

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