Is tan^2(θ)=(1-cos^2(θ))/(cos^2(θ)) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (θ)

Express with sin, cos

tan^{2} (θ)

Use the basic trigonometric identity:

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

= (\frac{sin ( θ )}{cos ( θ )})^{2}

Simplify

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

tan (u) - 1 tan (u) + 1 = 1 - cot (u) 1 + cot (u)

cos (x) = sec (x) - sin (x) tan (x)

arccsc (x) = sin (x)

cot (θ) = \frac{sin ( θ )}{cos ( θ )}

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