What is e^{-ln(cos(θ))+c} ?

The solution to e^{-ln(cos(θ))+c} is e^csec(θ)

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simplify e^{-ln(cos(θ))+c}

simplify

e^{- l n (c o s (θ)) + c}

e^{c} sec (θ)

Solution steps

e^{- l n (c o s (θ)) + c}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

= e^{- l n (c o s (θ))} e^{c}

Simplify

e^{- l n (c o s (θ))} : cos^{- 1} (θ)

= cos^{- 1} (θ) e^{c}

Apply exponent rule:

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

cos^{- 1} (θ) = \frac{1}{cos ( θ )}

= e^{c} \frac{1}{cos ( θ )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot e ^{c}}{cos ( θ )}

Multiply:

1 \cdot e^{c} = e^{c}

= \frac{e ^{c}}{cos ( θ )}

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= e^{c} sec (θ)

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