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

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

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prove (sin^2(x))/(sec^2(x)-1)=cos^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) + 1 = 2

tan^{2} (t) sin^{2} (t) = tan^{2} (t) - sin^{2} (t)

8 sin^{2} (θ) + 9 cos^{2} (θ) = 8 + cos^{2} (θ)

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

sec^{2} (x) + tan^{2} (x) = 1 + 2 tan^{2} (x)

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