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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{csc ( x ) + cot ( x )}

Rewrite using trig identities

\frac{1}{csc ( x ) + cot ( x )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

csc (x) + cot (x)

= csc (x) - cot (x)

= csc (x) - cot (x)

= csc (x) - cot (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (θ) (cot (θ) + tan (θ)) = sec^{2} (θ)

p ro v e sec^{2} (- θ) - 1 = tan^{2} (θ)

p ro v e 2 tan (x) = \frac{cos ( x )}{csc ( x ) - 1} + \frac{cos ( x )}{csc ( x ) + 1}

p ro v e sin (- θ) sec (- θ) cot (- θ) = 1

p ro v e cot (x) sin (x) sec^{2} (x) = sec (x)

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