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

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

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prove (1-cos(2x))/(cos^2(x))=2tan^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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 )}{cos ^{2} ( x )}

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

2 tan (x) tan (x)

Simplify

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

2 tan (x) tan (x)

Apply exponent rule:

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

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

= 2 tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (x)

2 tan^{2} (x)

2 tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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})

cos (θ) = sin (θ + \frac{π}{2})

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