Is (tan^4(x)-1)/(tan^2(x)-1)=sec^2(x) ?

The answer to whether (tan^4(x)-1)/(tan^2(x)-1)=sec^2(x) is True

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prove (tan^4(x)-1)/(tan^2(x)-1)=sec^2(x)

prove

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1} = sec^{2} (x)

True

Solution steps

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1} = sec^{2} (x)

Manipulating left side

\frac{tan ^{4} ( x ) - 1}{tan ^{2} ( x ) - 1}

Simplify

\frac{- 1 + tan ^{4} ( x )}{- 1 + tan ^{2} ( x )} : tan^{2} (x) + 1

= 1 + tan^{2} (x)

Rewrite using trig identities

= sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (27 0^{\circ} - θ) = - sin (θ)

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

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

sec (t) = \frac{cot ( t ) + tan ( t )}{csc ( t )}

\frac{1 - cos ( 2 a )}{sin ( 2 a )} = tan (a)

Is (tan^4(x)-1)/(tan^2(x)-1)=sec^2(x) ?
The answer to whether (tan^4(x)-1)/(tan^2(x)-1)=sec^2(x) is True