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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Rewrite using trig identities

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

Rewrite using trig identities

sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + sec ( β )}{tan ( β ) + sin ( β )} = csc (β)

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

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

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

\frac{1 + tan ( θ )}{1 + cot ( θ )} = tan (θ)

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