Is cot^6(x)=cot^4(x)csc^2(x)-cot^4(x) ?

The answer to whether cot^6(x)=cot^4(x)csc^2(x)-cot^4(x) is True

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prove cot^6(x)=cot^4(x)csc^2(x)-cot^4(x)

prove

cot^{6} (x) = cot^{4} (x) csc^{2} (x) - cot^{4} (x)

True

Solution steps

cot^{6} (x) = cot^{4} (x) csc^{2} (x) - cot^{4} (x)

Manipulating right side

cot^{4} (x) csc^{2} (x) - cot^{4} (x)

Rewrite using trig identities

cot^{4} (x) csc^{2} (x) - cot^{4} (x)

Use the Pythagorean identity:

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

= cot^{4} (x) (1 + cot^{2} (x)) - cot^{4} (x)

Simplify

cot^{4} (x) (1 + cot^{2} (x)) - cot^{4} (x) : cot^{6} (x)

cot^{4} (x) (1 + cot^{2} (x)) - cot^{4} (x)

Expand

cot^{4} (x) (1 + cot^{2} (x)) : cot^{4} (x) + cot^{6} (x)

cot^{4} (x) (1 + cot^{2} (x))

Apply the distributive law:

a (b + c) = ab + a c

a = cot^{4} (x), b = 1, c = cot^{2} (x)

= cot^{4} (x) \cdot 1 + cot^{4} (x) cot^{2} (x)

= 1 \cdot cot^{4} (x) + cot^{4} (x) cot^{2} (x)

Simplify

1 \cdot cot^{4} (x) + cot^{4} (x) cot^{2} (x) : cot^{4} (x) + cot^{6} (x)

1 \cdot cot^{4} (x) + cot^{4} (x) cot^{2} (x)

1 \cdot cot^{4} (x) = cot^{4} (x)

1 \cdot cot^{4} (x)

Multiply:

1 \cdot cot^{4} (x) = cot^{4} (x)

= cot^{4} (x)

cot^{4} (x) cot^{2} (x) = cot^{6} (x)

cot^{4} (x) cot^{2} (x)

Apply exponent rule:

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

cot^{4} (x) cot^{2} (x) = cot^{4 + 2} (x)

= cot^{4 + 2} (x)

Add the numbers:

4 + 2 = 6

= cot^{6} (x)

= cot^{4} (x) + cot^{6} (x)

= cot^{4} (x) + cot^{6} (x)

= cot^{4} (x) + cot^{6} (x) - cot^{4} (x)

Add similar elements:

cot^{4} (x) - cot^{4} (x) = 0

= cot^{6} (x)

= cot^{6} (x)

= cot^{6} (x)

We showed that the two sides could take the same form

\Rightarrow True

Is cot^6(x)=cot^4(x)csc^2(x)-cot^4(x) ?
The answer to whether cot^6(x)=cot^4(x)csc^2(x)-cot^4(x) is True