Is 1/(csc(Y)-cot(Y))=csc(Y)+cot(Y) ?

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

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prove 1/(csc(Y)-cot(Y))=csc(Y)+cot(Y)

prove

\frac{1}{csc ( Y ) - cot ( Y )} = csc (Y) + cot (Y)

True

Solution steps

\frac{1}{csc ( Y ) - cot ( Y )} = csc (Y) + cot (Y)

Manipulating left side

\frac{1}{csc ( Y ) - cot ( Y )}

Rewrite using trig identities

\frac{1}{csc ( Y ) - cot ( Y )}

Use the Pythagorean identity:

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

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

= \frac{csc ^{2} ( Y ) - cot ^{2} ( Y )}{csc ( Y ) - cot ( Y )}

Simplify

\frac{csc ^{2} ( Y ) - cot ^{2} ( Y )}{csc ( Y ) - cot ( Y )} : csc (Y) + cot (Y)

\frac{csc ^{2} ( Y ) - cot ^{2} ( Y )}{csc ( Y ) - cot ( Y )}

Apply Difference of Two Squares Formula:

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

csc^{2} (Y) - cot^{2} (Y) = (csc (Y) + cot (Y)) (csc (Y) - cot (Y))

= \frac{( csc ( Y ) + cot ( Y ) ) ( csc ( Y ) - cot ( Y ) )}{csc ( Y ) - cot ( Y )}

Cancel the common factor:

csc (Y) - cot (Y)

= csc (Y) + cot (Y)

= csc (Y) + cot (Y)

= csc (Y) + cot (Y)

We showed that the two sides could take the same form

\Rightarrow True

cos (π + θ) + cos (π - θ) = - 2 cos (θ)

sin (cot (x) + tan (x)) = sec (x)

sin^{2} (2 x) = 4 sin^{2} (x) \cdot cos^{2} (x)

\frac{sec ( x ) csc ( x )}{2} = csc (2 x)

- sec^{2} (x) = - tan^{2} (x) - 1

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