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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Rewrite using trig identities

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

Manipulating right side

tan (x) sec (x)

Express with sin, cos

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

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) - cos^{3} (θ) = sin^{2} (θ) cos (θ)

2 sin (θ) cos (θ) = tan (2 θ) cos (2 θ)

cos (θ) sec (θ) - cos^{2} (θ) = sin^{2} (θ)

sin (12 x) = 2 sin (6 x) cos (6 x)

\frac{sin ( t )}{csc ( t )} = 1 - \frac{cos ( t )}{sec ( t )}

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