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

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

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prove 1/(cos^2(x)2tan(x))=csc(2x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{cos ^{2} ( x ) 2 tan ( x )}

Express with sin, cos

\frac{1}{2 cos ^{2} ( x ) tan ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

2 cos^{2} (x) \frac{sin ( x )}{cos ( x )} : 2 sin (x) cos (x)

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 2 sin (x) cos (x)

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

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

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

Manipulating right 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 )}

We showed that the two sides could take the same form

\Rightarrow True

(1 - cos (θ)) (1 + sec (θ)) = sin (θ) tan (θ)

tan (y) - \frac{sec ^{2} ( y )}{tan ( y )} = - cot (y)

(csc (x) - cot (x)) (sec (x) 1) = tan (x)

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

tan (2 t) = 2 sin (t) cos (t) sec (2 t)

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