What is e^{-ln|1-sin(t)cos(t)|} ?

The solution to e^{-ln|1-sin(t)cos(t)|} is 1/(|-sin(t)cos(t)+1|)

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simplify e^{-ln|1-sin(t)cos(t)|}

simplify

e^{- l n ∣ 1 - s i n (t) c o s (t) ∣}

\frac{1}{∣ - sin ( t ) cos ( t ) + 1 ∣}

Solution steps

e^{- l n ∣ 1 - s i n (t) c o s (t) ∣}

Apply exponent rule:

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

= \frac{1}{e ^{l n ∣ 1 - s i n (t) c o s (t) ∣}}

e^{l n ∣ 1 - s i n (t) c o s (t) ∣} = ∣ 1 - sin (t) cos (t) ∣

= \frac{1}{∣ 1 - sin ( t ) cos ( t ) ∣}

16 a^{6} b^{6}

7 x^{2} (x - 2)^{2}

1 6^{- 2}

2 x^{2} (3 + x^{3})^{2}

What is e^{-ln|1-sin(t)cos(t)|} ?
The solution to e^{-ln|1-sin(t)cos(t)|} is 1/(|-sin(t)cos(t)+1|)