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^{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

1 + sin (x) cot (x) = 1 + cos (x)

\frac{cos ( x )}{sin ( x ) cot ( x )} = 1

sin^{2} (θ) cos (θ) sec (θ) = 1 - cos^{2} (θ)

sin (x) = \frac{cos ( x )}{cot ( x )}

(1 + sin (z)) (1 - sin (z)) = \frac{1}{sec ^{2} ( z )}

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