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

sec^{4} (x) = \frac{1}{cos ^{4} ( x )}

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

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

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

\frac{1}{tan ( x )} = cot (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