Is 2csc(2x)= 1/(sin(x)cos(x)) ?

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

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prove 2csc(2x)= 1/(sin(x)cos(x))

prove

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

True

Solution steps

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

Manipulating left side

2 csc (2 x)

Express with sin, cos

2 csc (2 x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Divide the numbers:

\frac{2}{2} = 1

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e (1 + csc (θ)) (1 - sin (θ)) = csc (θ) - sin (θ)

p ro v e \frac{1 - tan ( x )}{1 - cot ( x )} = - tan (x)

p ro v e \frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

p ro v e 1 + cot^{2} (- x) = csc^{2} (x)

p ro v e cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

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