Is cot(2x)+csc(2x)=cot(x) ?

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

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prove cot(2x)+csc(2x)=cot(x)

prove

cot (2 x) + csc (2 x) = cot (x)

True

Solution steps

cot (2 x) + csc (2 x) = cot (x)

Manipulating left side

cot (2 x) + csc (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 )}

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

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

Simplify

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

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

1 + 2 cos^{2} (x) - 1 = 2 cos^{2} (x)

1 + 2 cos^{2} (x) - 1

Group like terms

= 2 cos^{2} (x) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (x)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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