Is csc^4(x)-cot^4(x)=1+2cot^2(x) ?

The answer to whether csc^4(x)-cot^4(x)=1+2cot^2(x) is True

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prove csc^4(x)-cot^4(x)=1+2cot^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

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

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

Rewrite

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

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

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

Apply exponent rule:

a^{b c} = (a^{b})^{c}

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

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

Apply exponent rule:

a^{b c} = (a^{b})^{c}

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

csc^{2} (x) - cot^{2} (x) : (csc (x) + cot (x)) (csc (x) - cot (x))

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

Apply Difference of Two Squares Formula:

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

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

= (csc (x) + cot (x)) (csc (x) - cot (x))

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

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

Rewrite using trig identities

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

Expand

(csc (x) + cot (x)) (csc (x) - cot (x)) : csc^{2} (x) - cot^{2} (x)

(csc (x) + cot (x)) (csc (x) - cot (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = csc (x), b = cot (x)

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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Is csc^4(x)-cot^4(x)=1+2cot^2(x) ?
The answer to whether csc^4(x)-cot^4(x)=1+2cot^2(x) is True