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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (t)

Express with sin, cos

tan^{2} (t)

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{sin ( t )}{cos ( t )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( t )}{cos ^{2} ( t )}

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(2 sin (x) cos (x))^{4} = sin^{4} (2 x)

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

csc (θ) + sec (θ) = tan (θ) + 1

\frac{cos ( t ) + 1}{sin ( t ) + sec ( t )} = \frac{tan ( t )}{sec ( t ) - 1}

\frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )} = cos (2 θ)

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