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

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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

cos (4 x) = cos (3 x + x)

cot (θ) \cdot sec (θ) \cdot sin (θ) = 1

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

cos (2 0^{\circ}) = cos (θ) cos (θ) - sin (θ) sin (θ)

cos (θ) = \frac{4}{5}

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