Is csc(2θ)+cot(2θ)=cot(θ) ?

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

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prove csc(2θ)+cot(2θ)=cot(θ)

prove

csc (2 θ) + cot (2 θ) = cot (θ)

True

Solution steps

csc (2 θ) + cot (2 θ) = cot (θ)

Manipulating left side

csc (2 θ) + cot (2 θ)

Express with sin, cos

cot (2 θ) + csc (2 θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

= \frac{cos ( 2 θ ) + 1}{sin ( 2 θ )}

= \frac{cos ( 2 θ ) + 1}{sin ( 2 θ )}

= \frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

Rewrite using trig identities

\frac{1 + cos ( 2 θ )}{sin ( 2 θ )}

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

1 + 2 cos^{2} (θ) - 1 = 2 cos^{2} (θ)

1 + 2 cos^{2} (θ) - 1

Group like terms

= 2 cos^{2} (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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