Is csc^4(θ)-cot^4(θ)=2csc^2(θ)-1 ?

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

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prove csc^4(θ)-cot^4(θ)=2csc^2(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

csc^{4} (θ) - cot^{4} (θ)

Factor

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

- cot^{4} (θ) + csc^{4} (θ)

Rewrite

csc^{4} (θ) - cot^{4} (θ)

(csc^{2} (θ))^{2} - (cot^{2} (θ))^{2}

csc^{4} (θ) - cot^{4} (θ)

Apply exponent rule:

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

cot^{4} (θ) = (cot^{2} (θ))^{2}

= csc^{4} (θ) - (cot^{2} (θ))^{2}

Apply exponent rule:

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

csc^{4} (θ) = (csc^{2} (θ))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

csc^{2} (θ) - cot^{2} (θ)

Apply Difference of Two Squares Formula:

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

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

= (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

csc^{2} (θ) - 1 + csc^{2} (θ) : 2 csc^{2} (θ) - 1

csc^{2} (θ) - 1 + csc^{2} (θ)

Group like terms

= csc^{2} (θ) + csc^{2} (θ) - 1

Add similar elements:

csc^{2} (θ) + csc^{2} (θ) = 2 csc^{2} (θ)

= 2 csc^{2} (θ) - 1

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

Expand

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

(csc (θ) + cot (θ)) (csc (θ) - cot (θ))

Apply Difference of Two Squares Formula:

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

a = csc (θ), b = cot (θ)

= csc^{2} (θ) - cot^{2} (θ)

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

Use the Pythagorean identity:

csc^{2} (θ) = 1 + cot^{2} (θ)

csc^{2} (θ) - cot^{2} (θ) = 1

= (- 1 + 2 csc^{2} (θ)) \cdot 1

Simplify

= - 1 + 2 csc^{2} (θ)

= - 1 + 2 csc^{2} (θ)

= 2 csc^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

csc (x) - cos (x) cot (x) = sin (x)

\frac{csc ^{2} ( θ )}{cot ( θ )} = csc (θ) sec (θ)

\frac{cot ^{3} ( t )}{csc ( t )} = cos (t) (csc^{2} (t) - 1)

sin (2 x) - tan (x) = tan (x) cos (2 x)

\frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

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