Is (sec^2(θ)-1)/(tan(θ))=tan(θ) ?

The answer to whether (sec^2(θ)-1)/(tan(θ))=tan(θ) is True

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prove (sec^2(θ)-1)/(tan(θ))=tan(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ^{2} ( θ ) - 1}{tan ( θ )}

Rewrite using trig identities

\frac{sec ^{2} ( θ ) - 1}{tan ( θ )}

Use the Pythagorean identity:

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

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

= \frac{tan ^{2} ( θ )}{tan ( θ )}

Cancel the common factor:

tan (θ)

= tan (θ)

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( β )}{sec ( β )} = sin (β)

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

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

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

(1 - cos (x)) (1 + sec (x)) = - 1

Is (sec^2(θ)-1)/(tan(θ))=tan(θ) ?
The answer to whether (sec^2(θ)-1)/(tan(θ))=tan(θ) is True