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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (θ) + sec (θ) tan (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{1}{1 - sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{tan ^{2} ( θ )} - cos (θ) = cos^{2} (θ)

2 sin (x) = tan (x)

1 \cdot sin (t) + (- 1) \cdot cos (t - \frac{π}{2}) = 0

\frac{cot ( x )}{sec ( x ) - tan ( x )} = csc (x) + 1

csc (x) - cot (x) = \frac{1}{( csc ( x ) + cot ( x ) )}

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