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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (2 x) + cot (2 x)

Express with sin, cos

cot (2 x) + csc (2 x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

- cos (2 t) sin (2 t) + sin (2 t) cos (2 t) + 0 = 0

1 + sin (6 x) = (sin (3 x) + cos (3 x))^{2}

\frac{tan ( θ ) sin ( θ )}{sec ( θ ) - 1} = 1 + cos (θ)

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