What is the exact value of ${sec}^{- 1} (\sqrt{2}) and {csc}^{- 1} (2)$ $sec^-1 (sqrt2) and csc^-1 (2)$ ?

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What is the exact value of ${sec}^{- 1} (\sqrt{2}) and {csc}^{- 1} (2)$ $sec^-1 (sqrt2) and csc^-1 (2)$ ?

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Answer 1 · 2018-03-28T10:39:28

${sec}^{- 1} (\sqrt{2}) = \frac{π}{4}$ $sec^-1(sqrt2)=pi/4$
${csc}^{- 1} (2) = \frac{π}{6}$ $csc^-1(2)=pi/6$

Explanation:

Note:
$(1) {sec}^{- 1} x = {cos}^{- 1} (\frac{1}{x})$ $color(red)((1)sec^-1x=cos^-1(1/x)$

$(2) {csc}^{- 1} x = {sin}^{- 1} (\frac{1}{x})$ $color(red)((2)csc^-1x=sin^-1(1/x)$

$(3) {cos}^{- 1} (cos θ) = θ, w h e r e, θ \in [0, π]$ $color(red)((3)cos^-1(costheta))=color(red)(theta,where,theta in [0,pi]$

$(4) {sin}^{- 1} (sin θ) = θ, w h e r e, θ \in [- \frac{π}{2}, \frac{π}{2}]$ $color(red)((4)sin^-1(sintheta))=color(red)(theta,where,theta in [-pi/2,pi/2]$

Here,

$sec^-1(sqrt2)=cos^-1(1/sqrt2)...toApply (1)$

$sec^-1(sqrt2)=cos^-1(cos(pi/4))$

$sec^-1(sqrt2)=pi/4in[0,pi].............toApply(3)$

Now,

$csc^-1(2)=sin^-1(1/2).....toApply(2)$

$csc^-1(2)=sin^-1(sin(pi/6))$

$csc^-1(2)=pi/6in[-pi/2,pi/2]..........toApply(4)$

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What is the exact value of ${sec}^{- 1} (\sqrt{2}) and {csc}^{- 1} (2)$ $sec^-1 (sqrt2) and csc^-1 (2)$ ?

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Explanation:

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What is the exact value of ${sec}^{- 1} (\sqrt{2}) and {csc}^{- 1} (2)$ $sec^-1 (sqrt2) and csc^-1 (2)$ ?