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

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

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prove (cos(2x))/(1-sin^2(x))=2-sec^2(x)

prove

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

True

Solution steps

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

Manipulating right side

2 - sec^{2} (x)

Express with sin, cos

2 - sec^{2} (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

cot (\frac{y}{2}) = \frac{sin ( y )}{1 - cos ( y )}

cot (- θ) \cdot cos (- θ) + sin (- θ) = - csc (θ)

cot^{2} (x) - csc^{2} (x) = 1

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

tan^{2} (2 π) = tan (2 π)

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