Is (cos(2θ))/(cos^2(θ))=1-tan^2(θ) ?

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

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prove (cos(2θ))/(cos^2(θ))=1-tan^2(θ)

prove

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

True

Solution steps

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

Manipulating right side

1 - tan^{2} (θ)

Express with sin, cos

1 - tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

1 - (\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= \frac{cos ( 2 θ )}{cos ^{2} ( θ )}

= \frac{cos ( 2 θ )}{cos ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

(tan (θ + 1)) = sec (θ)

\frac{sin ( x ) csc ( x )}{cos ( x )} = sec (x)

tan (x) + sec (x) = 1

- sinh^{2} (x) + cosh^{2} (x) = 1

sin^{2} (x) + cos^{2} (x) - 1 = 0

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