Question #0fe54

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T15:46:20

The answer is $= 2$ $=2$

If $cot x = \frac{5}{12}$ $cotx=5/12$

$tan x = \frac{12}{5}$ $tanx=12/5$

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$

Dividing by ${cos}^{2} x$ $cos^2x$

$1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$

${sec}^{2} x = 1 + \frac{12^{2}}{5^{2}} = 1 + \frac{144}{25} = \frac{169}{25}$ $sec^2x=1+12^2/5^2=1+144/25=169/25$

$sec x = \frac{13}{5}$ $secx=13/5$

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

$cos x = \frac{5}{13}$ $cosx=5/13$

$sin x = \sqrt{1 - {cos}^{2} x} = \sqrt{1 - \frac{25}{169}}$ $sinx=sqrt(1-cos^2x)=sqrt(1-25/169)$

$= \frac{\sqrt{144}}{169} = \frac{12}{13}$ $=sqrt144/169=12/13$

$csc x = \frac{1}{sin x} = \frac{13}{12}$ $cscx=1/sinx=13/12$

And finally,

$sin x + csc x = \frac{12}{13} + \frac{13}{12} = 2$ $sinx+cscx=12/13+13/12=2$