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

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

p ro v e cos (θ) (cot (θ) + tan (θ)) = csc (θ)

p ro v e sin (- θ) = - sin (θ)

p ro v e - sin (x) = sin (- x)

p ro v e \frac{sin ( y ) + tan ( y )}{1 + sec ( y )} = sin (y)

p ro v e cos^{2} (2 x) - sin^{2} (2 x) = cos (4 x)

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