Is 1/(cos(θ))-cos(θ)=sin(θ)tan(θ) ?

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

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prove 1/(cos(θ))-cos(θ)=sin(θ)tan(θ)

prove

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

True

Solution steps

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

Manipulating right side

sin (θ) tan (θ)

Express with sin, cos

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

= \frac{1 - cos ^{2} ( θ )}{cos ( θ )}

Expand

\frac{1 - cos ^{2} ( θ )}{cos ( θ )} : \frac{1}{cos ( θ )} - cos (θ)

= - cos (θ) + \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )} - cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

cot^{2} (x) (1 - cos^{2} (x)) = cos^{2} (x)

sec^{2} (θ) = 1 + tan^{2} (θ)

1 + tan^{2} (3 0^{\circ}) = sec^{2} (3 0^{\circ})

(sin (x))^{2} - (sec (x)) (sin (x))^{3} = - 1

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