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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Rewrite using trig identities

= \frac{1}{1 - sin ( x )}

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{csc ( x )}{cos ( x )} = tan (x) + cot (x)

\frac{1 + cos ( 2 θ )}{sin ( 2 θ )} = cot (θ)

cot (x) \cdot sin (x) = cos (x)

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )} = 1

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