Is tan^3(θ)=tan(θ)sec^2(θ)-tan(θ) ?

The answer to whether tan^3(θ)=tan(θ)sec^2(θ)-tan(θ) is True

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prove tan^3(θ)=tan(θ)sec^2(θ)-tan(θ)

prove

tan^{3} (θ) = tan (θ) sec^{2} (θ) - tan (θ)

True

Solution steps

tan^{3} (θ) = tan (θ) sec^{2} (θ) - tan (θ)

Manipulating right side

tan (θ) sec^{2} (θ) - tan (θ)

Factor

- tan (θ) + sec^{2} (θ) tan (θ) : tan (θ) (- 1 + sec^{2} (θ))

= (- 1 + sec^{2} (θ)) tan (θ)

Rewrite using trig identities

= tan^{3} (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (θ) + sin (θ) = 1

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

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

sin (θ - \frac{3 π}{2}) = cos (θ)

cos^{2} (x) + 2 sin (x) = 2

Is tan^3(θ)=tan(θ)sec^2(θ)-tan(θ) ?
The answer to whether tan^3(θ)=tan(θ)sec^2(θ)-tan(θ) is True