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

\frac{1}{sec ( θ ) + tan ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

\frac{sec ^{2} ( θ ) - tan ^{2} ( θ )}{sec ( θ ) + tan ( θ )} : sec (θ) - tan (θ)

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (θ) - tan^{2} (θ) = (sec (θ) + tan (θ)) (sec (θ) - tan (θ))

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

Cancel the common factor:

sec (θ) + tan (θ)

= sec (θ) - tan (θ)

= sec (θ) - tan (θ)

= 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

2 cos^{3} (x) - cos (x) = \frac{cos ^{2} ( x ) - sin ^{2} ( x )}{sec ( x )}

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

\frac{cos ( 5 x ) - cos ( 3 x )}{sin ( 5 x ) + sin ( 3 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