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

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

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

prove

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

True

Solution steps

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

Manipulating right 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

cos (π - θ) + sin (\frac{π}{2} + θ) = 0

sin (x) - csc (x) = - cos (x) cot (x)

tan (y) + cot (y) = sec (y) csc (y)

\frac{1}{sin ( x )} = csc (x)

\frac{csc ( θ )}{sec ( θ )} = cot (θ)

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