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

sin (u) (csc (u) - sin (u)) = cos^{2} (u)

cos^{2} (θ) - sin^{2} (θ) = 1

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

\frac{1 - tan ^{4} ( θ )}{sec ^{2} ( θ )} = 1 - tan^{2} (θ)

(1 + csc (θ)) (1 - sin (θ)) = csc (θ) - sin (θ)

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