How do you prove ${sec}^{- 1} x + {csc}^{- 1} x = \frac{π}{2}$ $sec^-1x+csc^-1x=pi/2$ ?

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Answer 1 · 2016-10-20T16:28:50

Please see the explanation.

Prove:

${sec}^{- 1} (x) + {csc}^{- 1} (x) = \frac{π}{2}$ $sec^-1(x) + csc^-1(x) = pi/2$

Use the identity ${csc}^{- 1} (x) = \frac{π}{2} - {sec}^{- 1} (x)$ $csc^-1(x) = pi/2 - sec^-1(x)$ :

${sec}^{- 1} (x) + \frac{π}{2} - {sec}^{- 1} (x) = \frac{π}{2}$ $sec^-1(x) + pi/2 - sec^-1(x) = pi/2$

$\frac{π}{2} = \frac{π}{2}$ $pi/2 = pi/2$

Q.E.D.

Answer 2 · 2016-10-20T17:15:46

use the fact that csc is the complementary function of sec.

let
${sec}^{- 1} x = y$ $sec^-1x=y$

$\Rightarrow x = sec y$ $=>x=secy$

$\Rightarrow x = csc (\frac{π}{2} - y)$ $=>x=csc(pi/2-y)$
$\Rightarrow {csc}^{- 1} x = \frac{π}{2} - y$ $=>csc^-1x=pi/2-y$

substituting back for y

${csc}^{- 1} x = \frac{π}{2} - {sec}^{- 1} x$ $csc^-1x=pi/2-sec^-1x$

hence ${sec}^{- 1} x + {csc}^{- 1} x = \frac{π}{2}$ $sec^-1x+csc^-1x=pi/2$

as required.

How do you prove sec^-1x+csc^-1x=pi/2sec−1x+csc−1x=π2sec^-1x+csc^-1x=pi/2?