Is tan^5(x)=tan^3(x)sec^2(x)-tan^3(x) ?

The answer to whether tan^5(x)=tan^3(x)sec^2(x)-tan^3(x) is True

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prove tan^5(x)=tan^3(x)sec^2(x)-tan^3(x)

prove

tan^{5} (x) = tan^{3} (x) sec^{2} (x) - tan^{3} (x)

True

Solution steps

tan^{5} (x) = tan^{3} (x) sec^{2} (x) - tan^{3} (x)

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= tan^{3} (x) (tan^{2} (x) + 1) - tan^{3} (x)

Simplify

tan^{3} (x) (tan^{2} (x) + 1) - tan^{3} (x) : tan^{5} (x)

tan^{3} (x) (tan^{2} (x) + 1) - tan^{3} (x)

Expand

tan^{3} (x) (tan^{2} (x) + 1) : tan^{5} (x) + tan^{3} (x)

tan^{3} (x) (tan^{2} (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = tan^{3} (x), b = tan^{2} (x), c = 1

= tan^{3} (x) tan^{2} (x) + tan^{3} (x) \cdot 1

= tan^{3} (x) tan^{2} (x) + 1 \cdot tan^{3} (x)

Simplify

tan^{3} (x) tan^{2} (x) + 1 \cdot tan^{3} (x) : tan^{5} (x) + tan^{3} (x)

tan^{3} (x) tan^{2} (x) + 1 \cdot tan^{3} (x)

tan^{3} (x) tan^{2} (x) = tan^{5} (x)

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

Apply exponent rule:

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

tan^{3} (x) tan^{2} (x) = tan^{3 + 2} (x)

= tan^{3 + 2} (x)

Add the numbers:

3 + 2 = 5

= tan^{5} (x)

1 \cdot tan^{3} (x) = tan^{3} (x)

1 \cdot tan^{3} (x)

Multiply:

1 \cdot tan^{3} (x) = tan^{3} (x)

= tan^{3} (x)

= tan^{5} (x) + tan^{3} (x)

= tan^{5} (x) + tan^{3} (x)

= tan^{5} (x) + tan^{3} (x) - tan^{3} (x)

Add similar elements:

tan^{3} (x) - tan^{3} (x) = 0

= tan^{5} (x)

= tan^{5} (x)

= tan^{5} (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{cos ( θ )}{1 - sin ( θ )} = sec (θ) + tan (θ)

p ro v e tan (x) + cot (x) = 2 csc (2 x)

p ro v e sin (\frac{π}{2} - x) = cos (x)

p ro v e tan (2 x) = \frac{( 2 tan ( x ) )}{( 1 - tan ^{2} ( x ) )}

p ro v e cos^{2} (x) = \frac{1 + cos ( 2 x )}{2}

Is tan^5(x)=tan^3(x)sec^2(x)-tan^3(x) ?
The answer to whether tan^5(x)=tan^3(x)sec^2(x)-tan^3(x) is True