Is (cot(2x))/(cos(2x))=csc(2x) ?

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

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prove (cot(2x))/(cos(2x))=csc(2x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{cot ( 2 x )}{cos ( 2 x )}

Express with sin, cos

\frac{cot ( 2 x )}{cos ( 2 x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

cos (2 x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( 2 x )}}

Simplify

\frac{1}{\frac{1}{c s c ( 2 x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc (2 x)

csc (2 x)

csc (2 x)

We showed that the two sides could take the same form

\Rightarrow True

(tan (θ) + 1) (tan (θ) - 1) = sec^{2} (θ) - 2

\frac{sec ( t ) - 1}{1 - cos ( t )} = sec (t)

tan (\frac{7 π}{6}) = \frac{3}{3}

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

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

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