How do you find the exact value of ${sec}^{- 1} (\sqrt{2})$ $sec^-1(sqrt2)$ ?

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How do you find the exact value of ${sec}^{- 1} (\sqrt{2})$ $sec^-1(sqrt2)$ ?

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Answer 1 · 2016-09-01T18:55:17

45degrees or $\frac{π}{4}$ $pi/4$

Explanation:

Draw a. Right angled isosceles triangle . The ratio of the sides 1:1: $\sqrt{2}$ $sqrt2$
Sec is hypotenuse over adjacent

Answer 2 · 2016-09-01T18:57:34

${sec}^{- 1} \sqrt{2} = \frac{π}{4}$ $sec^-1 sqrt2=pi/4$ .

Explanation:

The Defn. of ${sec}^{- 1}$ $sec^-1$ is :

${sec}^{- 1} x = θ, | x | . = 1 \Leftrightarrow sec θ = x, θ \in [0, π] - {\frac{π}{2}}$ $sec^-1 x=theta , |x|.=1 iff sec theta=x, theta in [0,pi]-{pi/2}$ .

Knowing that, $sec (\frac{π}{4}) = \sqrt{2}, \frac{π}{4} \in [0, 2 π] - {\frac{π}{2}}, {sec}^{- 1} \sqrt{2} = \frac{π}{4}$ $sec (pi/4)= sqrt2, pi/4 in [0,2pi]-{pi/2}, sec^-1 sqrt2=pi/4$ .

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How do you find the exact value of ${sec}^{- 1} (\sqrt{2})$ $sec^-1(sqrt2)$ ?

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How do you find the exact value of ${sec}^{- 1} (\sqrt{2})$ $sec^-1(sqrt2)$ ?