Is tan^8(x)=tan^6(x)sec^2(x)-tan^6(x) ?

The answer to whether tan^8(x)=tan^6(x)sec^2(x)-tan^6(x) is True

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prove tan^8(x)=tan^6(x)sec^2(x)-tan^6(x)

prove

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

True

Solution steps

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Manipulating right side

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Rewrite using trig identities

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Use the Pythagorean identity:

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

= tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

Simplify

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x) : tan^{8} (x)

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

Expand

tan^{6} (x) (tan^{2} (x) + 1) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) (tan^{2} (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = tan^{6} (x), b = tan^{2} (x), c = 1

= tan^{6} (x) tan^{2} (x) + tan^{6} (x) \cdot 1

= tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

Simplify

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

tan^{6} (x) tan^{2} (x) = tan^{8} (x)

tan^{6} (x) tan^{2} (x)

Apply exponent rule:

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

tan^{6} (x) tan^{2} (x) = tan^{6 + 2} (x)

= tan^{6 + 2} (x)

Add the numbers:

6 + 2 = 8

= tan^{8} (x)

1 \cdot tan^{6} (x) = tan^{6} (x)

1 \cdot tan^{6} (x)

Multiply:

1 \cdot tan^{6} (x) = tan^{6} (x)

= tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x) - tan^{6} (x)

Add similar elements:

tan^{6} (x) - tan^{6} (x) = 0

= tan^{8} (x)

= tan^{8} (x)

= tan^{8} (x)

We showed that the two sides could take the same form

\Rightarrow True

2 (sin (2 y) cos (2 y)) = sin (4 y)

csc^{2} (θ) \cdot (sin^{2} (θ)) = 1

tan (b) cos (b) + cot (b) cos (b) = csc (b)

1 + cot^{2} (z) = csc^{2} (z)

csc (x) - 1 = \frac{2 sin ( x )}{sin ( x )}

Is tan^8(x)=tan^6(x)sec^2(x)-tan^6(x) ?
The answer to whether tan^8(x)=tan^6(x)sec^2(x)-tan^6(x) is True