Is tan^3(x)=tan(x)sec^2(x)-tan(x) ?

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

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prove tan^3(x)=tan(x)sec^2(x)-tan(x)

prove

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

True

Solution steps

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

Manipulating right side

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

Factor

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

- tan (x) + sec^{2} (x) tan (x)

Factor out common term

tan (x)

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= tan^{2} (x) tan (x)

Simplify

tan^{2} (x) tan (x) : tan^{3} (x)

tan^{2} (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= tan^{3} (x)

= tan^{3} (x)

= tan^{3} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) (sec (θ) - cos (θ)) = sin^{2} (θ)

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

\frac{2 cot ( x )}{csc ^{2} ( x )} = sin (2 x)

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

tan (x) - cot (x) = - 2 cot (2 x)

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