Is (tan^2(θ))/(1-cos^2(θ))=sec^2(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

\frac{tan ^{2} ( θ )}{sin ^{2} ( θ )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

Simplify

\frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )} : \frac{1}{cos ^{2} ( θ )}

\frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sin ^{2} ( θ )}

(\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{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{sin ^{2} ( θ )}

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}