Is (sec(θ)-1)/(sec(θ)+1)=(tan(θ)-sin(θ))/(tan(θ)+sin(θ)) ?

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{tan ( θ ) - sin ( θ )}{tan ( θ ) + sin ( θ )}

Express with sin, cos

= \frac{1 - cos ( θ )}{1 + cos ( θ )}

Rewrite using trig identities

\frac{sec ( θ ) - 1}{sec ( θ ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

cos^{4} (a) - sin^{4} (a) + 1 = 2 cos^{2} (a)

(csc (θ)) (sec (θ)) - cot (θ) = tan (θ)

(cos (x)) (cot (x)) = csc (x) - sin (x)

\frac{sec ( - x )}{tan ( x ) + cot ( x )} = sin (x)

sin (3 a) + sin (a) = 4 sin (a) - 4 sin^{3} (a)

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