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

\frac{1 - sin ( x )}{1 - csc ( x )} = - sin (x)

1 + cos (10 y) = 2 cos^{2} (5 y)

cos (a) tan (a) = sin (a)

sin (\frac{π}{6}) = \frac{1}{2}

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