Is (csc(θ)-cot(θ))(csc(θ)+cot(θ))=1 ?

The answer to whether (csc(θ)-cot(θ))(csc(θ)+cot(θ))=1 is True

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prove (csc(θ)-cot(θ))(csc(θ)+cot(θ))=1

prove

(csc (θ) - cot (θ)) (csc (θ) + cot (θ)) = 1

True

Solution steps

(csc (θ) - cot (θ)) (csc (θ) + cot (θ)) = 1

Manipulating left side

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

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)

= 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

We showed that the two sides could take the same form

\Rightarrow True

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