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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

csc (t) - cot (t)

= csc (t) + cot (t)

= csc (t) + cot (t)

= csc (t) + cot (t)

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{5 π}{2} + x) = cos (x)

\frac{2}{cos ^{2} ( x ) - tan ^{2} ( x )} = 2 + tan^{2} (x)

cos (x) = sin (x + \frac{π}{6}) - sin (x - \frac{π}{6})

tan^{2} (\frac{π}{4}) = tan (\frac{π}{4})

1 - sin (x) cos (x) cot (x) = sin^{2} (x)

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