Is (1+cos(2x))/(1-cos(2x))=cot^2(x) ?

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

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prove (1+cos(2x))/(1-cos(2x))=cot^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Add the numbers:

1 + 1 = 2

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

Expand

1 - (1 - 2 sin^{2} (x)) : 2 sin^{2} (x)

1 - (1 - 2 sin^{2} (x))

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

- (1 - 2 sin^{2} (x))

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Factor

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

- 2 sin^{2} (x) + 2

Rewrite as

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

Factor out common term

2

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

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

Divide the numbers:

\frac{2}{2} = 1

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

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

cot (x) cot (x)

Simplify

cot (x) cot (x) : cot^{2} (x)

cot (x) cot (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (x) cot (x) = cot^{1 + 1} (x)

= cot^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cot^{2} (x)

cot^{2} (x)

cot^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (θ) = (sin (θ)) (sin (θ))

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

cos (4 t) = 1 - 8 sin^{2} (t) cos^{2} (t)

sin (10 5^{\circ}) = sin (6 0^{\circ} + 4 5^{\circ})

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