What is e^{-ln|cot(x+y)+y|} ?

The solution to e^{-ln|cot(x+y)+y|} is 1/(|y+cot(x+y)|)

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simplify e^{-ln|cot(x+y)+y|}

simplify

e^{- l n ∣ c o t (x + y) + y ∣}

\frac{1}{∣ y + cot ( x + y ) ∣}

Solution steps

e^{- l n ∣ c o t (x + y) + y ∣}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{l n ∣ c o t (x + y) + y ∣}}

e^{l n ∣ c o t (x + y) + y ∣} = ∣ cot (x + y) + y ∣

= \frac{1}{∣ cot ( x + y ) + y ∣}

225 \cdot 1 0^{- 6} \frac{1600}{π} \cdot \frac{1}{6}

(12 a c^{2} b) \cdot (12 a x c^{2})

- (a - 1)^{3} \cdot (a + 1)^{3}

e^{- 22495.8282}

(2 cos (x))^{2}

What is e^{-ln|cot(x+y)+y|} ?
The solution to e^{-ln|cot(x+y)+y|} is 1/(|y+cot(x+y)|)