Is 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ) ?

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

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prove 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{sec ( θ ) - tan ( θ )}

Rewrite using trig identities

= sec (θ) + tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (4 a) = 4 sin (a) cos (a) cos (2 a)

3 cos^{2} (x) - 2 = 1 - 3 sin^{2} (x)

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

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

sin^{2} (x) sec (x) csc (x) = tan (x)

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