Is tan^2(θ)= 1/(cos^2(θ))-1 ?

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

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prove tan^2(θ)= 1/(cos^2(θ))-1

prove

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

True

Solution steps

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

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 )}

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

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

= \frac{1}{cos ^{2} ( θ )} - 1

= - 1 + \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )} - 1

We showed that the two sides could take the same form

\Rightarrow True

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\frac{1 + tan ( u )}{1 + cot ( u )} = tan (u)

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

3 sin^{2} (2 x) = 12 cos^{2} (x) sin^{2} (x)

cos (θ) = sin (θ) cot (θ)

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