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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x)

Express with sin, cos

tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) (csc (θ) - sin (θ)) = cos (θ)

cot^{2} (x) sec^{2} (x) = csc^{2} (x)

sin (10 θ) = 2 sin (5 θ) cos (5 θ)

\frac{cos ( θ ) sec ( θ )}{tan ( θ )} = cot (θ)

\frac{cos ( θ ) - 1}{sin ( θ )} = cot (θ) - csc (θ)

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