Is cot^2(x)=(csc(x)-1)(csc(x)+1) ?

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

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prove cot^2(x)=(csc(x)-1)(csc(x)+1)

prove

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

True

Solution steps

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

Manipulating left side

cot^{2} (x)

Use the Pythagorean identity:

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

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

= csc^{2} (x) - 1

Factor

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

csc^{2} (x) - 1

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

csc^{2} (x) - 1 = (csc (x) + 1) (csc (x) - 1)

= (csc (x) + 1) (csc (x) - 1)

= (csc (x) + 1) (csc (x) - 1)

We showed that the two sides could take the same form

\Rightarrow True

sin (θ) - sin (θ) cos^{2} (θ) = sin^{3} (θ)

sin (x + \frac{π}{2}) = cos (x)

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

csc^{4} (x) - 2 csc^{2} (x) + 1 = cot^{4} (x)

\frac{cos ( x )}{1 - sin ( x )} + \frac{1 - sin ( x )}{cos ( x )} = 2 sec (x)

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