Is cot(θ)=((1+cos(2θ)))/(sin(2θ)) ?

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

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prove cot(θ)=((1+cos(2θ)))/(sin(2θ))

prove

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

True

Solution steps

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

Manipulating right side

\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

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

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

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )} = sec (A)

sin (x - \frac{π}{6}) + cos (x) = sin (x + π)

tan^{2} (2 3^{\circ}) - sec^{2} (2 3^{\circ}) = - 1

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