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

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

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

prove

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

True

Solution steps

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

Manipulating right 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 )}

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{sec ( x )}{tan ( x )} = csc (x)

\frac{1 - sin ( v )}{cos ( v )} + \frac{cos ( v )}{1 - sin ( v )} = 2 sec (v)

sec (θ) - tan (θ) = \frac{1}{sec ( θ ) + tan ( θ )}

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