Is 1/(csc(θ)-cot(θ))=csc(θ)+cot(θ) ?

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

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prove 1/(csc(θ)-cot(θ))=csc(θ)+cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{csc ( θ ) - cot ( θ )}

Rewrite using trig identities

\frac{1}{csc ( θ ) - cot ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

csc (θ) - cot (θ)

= csc (θ) + cot (θ)

= csc (θ) + cot (θ)

= csc (θ) + cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

sin (θ) - sin (θ - 2 π) = 0

- sin (2 x) = - 2 sin (x) cos (x)

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