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Popular >

arcsech(1)

Solution

arcsech (1)

Solution

0

Solution steps

arcsech (1)

Use the Hyperbolic identity:

arcsech (x) = ln (\frac{1}{x} + \frac{1}{x ^{2}} - 1)

= ln (\frac{1}{1} + \frac{1}{1 ^{2}} - 1)

ln (\frac{1}{1} + \frac{1}{1 ^{2}} - 1) = 0

ln (\frac{1}{1} + \frac{1}{1 ^{2}} - 1)

Simplify

\frac{1}{1} + \frac{1}{1 ^{2}} - 1 : \frac{1}{1} - 1 + 1

\frac{1}{1} + \frac{1}{1 ^{2}} - 1

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{1} + \frac{1}{1} - 1

Apply rule

\frac{a}{a} = 1

\frac{1}{1} = 1

= 1 + \frac{1}{1} - 1

= ln (1 + \frac{1}{1} - 1)

\frac{1}{1} - 1 = 0

\frac{1}{1} - 1

Apply rule

\frac{a}{1} = a

\frac{1}{1} = 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

Apply rule

0 = 0

= 0

= ln (1 + 0)

Add the numbers:

1 + 0 = 1

= ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

= 0

Popular Examples

arccos (- \frac{6}{10})

tan (32. 5^{\circ})

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arcsin (\frac{6}{13})

tan (π) 10

Frequently Asked Questions (FAQ)

What is the value of arcsech(1) ?
The value of arcsech(1) is 0