Is (csc(θ)+1)(csc(θ)-1)=cot^2(θ) ?

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

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prove (csc(θ)+1)(csc(θ)-1)=cot^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

(csc (θ) + 1) (csc (θ) - 1)

Expand

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

(csc (θ) + 1) (csc (θ) - 1)

Apply Difference of Two Squares Formula:

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

a = csc (θ), b = 1

= csc^{2} (θ) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= csc^{2} (θ) - 1

= csc^{2} (θ) - 1

Use the Pythagorean identity:

csc^{2} (θ) = 1 + cot^{2} (θ)

csc^{2} (θ) - 1 = cot^{2} (θ)

= cot^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) (cot (θ) + tan (θ)) = csc (θ)

sin (- θ) = - sin (θ)

sin (2 t) - cot (t) = - cot (t) cos (2 t)

- sin (x) = sin (- x)

sec (θ) cot (θ) sin (θ) = 1

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