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 left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

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

- tan^{2} (x) tan (x) - 1 \cdot tan (x)

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)

1 \cdot tan (x) = tan (x)

1 \cdot tan (x)

Multiply:

1 \cdot tan (x) = tan (x)

= tan (x)

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

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

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

Add similar elements:

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

= - tan (x)

= - tan (x)

= - tan (x)

Manipulating right side

tan (- x)

Use the negative angle identity:

tan (- x) = - tan (x)

= - tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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