Is (cot(X)-csc(X))(cot(X)+csc(X))=-1 ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

(cot (X) - csc (X)) (cot (X) + csc (X))

Expand

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

(cot (X) + csc (X)) (cot (X) - csc (X))

Apply Difference of Two Squares Formula:

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

a = cot (X), b = csc (X)

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

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

Use the Pythagorean identity:

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

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

= - 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{2 sin ( 2 θ ) cos ( θ )}{2 sin ( 2 θ )} = cos (θ)

p ro v e - sin (2 x) + cos (2 x) = 1

p ro v e sin (x) (sec (x) tan (x)) = tan^{2} (x)

p ro v e tan (θ) + cot (θ) = \frac{1}{sin ( θ ) cos ( θ )}

p ro v e cot^{2} (x) + csc^{2} (x) = 1 + cot^{2} (x)

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