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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Factor

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

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

Apply exponent rule:

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

csc^{4} (x) = csc^{2} (x) csc^{2} (x)

= - csc^{2} (x) + csc^{2} (x) csc^{2} (x)

Factor out common term

csc^{2} (x)

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Expand

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

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

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

Simplify

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

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

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

1 \cdot cot^{2} (x)

Multiply:

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

= cot^{2} (x)

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

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

Apply exponent rule:

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

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

= cot^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= cot^{4} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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