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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

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

Rewrite using trig identities

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

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

= cot (x) - csc (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (θ + 3 0^{\circ}) - sin (θ + 6 0^{\circ}) = - sin (θ)

\frac{tan ^{2} ( x )}{sec ( x )} = sin (tan (x))

sin^{2} (0) = 2 - 2 cos (0)

cos (\frac{3 π}{8}) = \frac{2 - 2}{2}

tan (a) \cdot cot (a) = sin^{2} (a) + cos^{2} (a)

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