Is sec(θ)-tan(θ)= 1/(sec(θ)+tan(θ)) ?

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

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prove sec(θ)-tan(θ)= 1/(sec(θ)+tan(θ))

prove

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

True

Solution steps

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

Manipulating right 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 (x) = \frac{1 + csc ( x )}{cot ( x ) + cos ( x )}

9 sec^{2} (θ) - 5 tan^{2} (θ) = 5 + 4 sec^{2} (θ)

cot (2 θ) = cot (θ) - \frac{1}{2} sec (θ) csc (θ)

(csc (x) - cot (x)) (csc (x) + cot (x)) = 1

(tan (x)) (sin (x)) = sec (x) - cos (x)

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