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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - tan^{2} (x)

Express with sin, cos

1 - tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (- θ) + cos^{2} (- θ) = 1

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

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

2 cot (θ) cot (2 θ) = cot^{2} (θ) - 1

csc (u) - cot (u) = \frac{sin ( u )}{1 + cos ( u )}

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