We have,sech−1(51)We know that, sech−1x=log(x1+x211)Now,sec−151 =log⎝⎛51 1+(51 ) 21−1 ⎠⎞ =log(5+25−1 ) =log(5+24 ) Hence this is the answer.

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h^{-1}({1}{5})=

Question

$sec h^{- 1} (\frac{1}{5})$ =

lo g (24 + 5)

lo g 5 + 27

lo g 26 + 5

lo g 27 + 5

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Text solutionVerified

We have,

sec h^{- 1} (\frac{1}{5})

We know that,

sec h^{- 1} x = lo g (\frac{1}{x} + \frac{1}{x ^{2}} 1)

Now,

sec^{- 1} \frac{1}{5} = lo g ⎝ ⎛ \frac{1}{\frac{1}{5}} + \frac{1}{( \frac{1}{5} ) ^{2}} - 1 ⎠ ⎞

= lo g (5 + 25 - 1)

= lo g (5 + 24)

Hence this is the answer.

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