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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Rewrite using trig identities

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

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (x) cot^{2} (x) = cos^{2} (x)

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

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

sin (a + b) sin (a - b) = sin^{2} (b) - sin^{2} (a)

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