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

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

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

prove

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

True

Solution steps

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

Manipulating right 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

4 sin^{2} (x) = 3 - 4 sin (x)

sec^{4} (x) - tan^{4} (x) = 1 + 2 tan^{4} (x)

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

cos (\frac{1}{sin ( b )}) \frac{sin ( b )}{cos ( b )} = 1

8 tan^{2} (y) (csc^{2} (y) - 1) = 8

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