Question #18b64

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Question #18b64

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Answer 1 · 2017-03-18T14:26:09

13.23

Explanation:

$tan x + \frac{1}{tan x} = 2$ $tan x + 1/tan x = 2$
${tan}^{2} x + 1 = 2 tan x$ $tan^2 x + 1 = 2tan x$
${tan}^{2} x - 2 tan x + 1 = 0$ $tan^2 x - 2tan x + 1 = 0$
Solve this quadratic equation for tan x.
Since a+ b + c = 0, use shortcut. There are 2 real roots:
tan x = 1 and $tan x = \frac{c}{a} = \frac{1}{2}$ $tan x = c/a = 1/2$ .
a. tan x = 1 --> $x = \frac{π}{4}$ $x = pi/4$
$10 x - \to \frac{10 π}{4} - \to \frac{5 π}{2} - \to \frac{π}{2}$ $10x --> (10pi)/4 --> (5pi)/2 --> pi/2$
$tan 10 x + cot 10 x = tan (\frac{π}{2}) + cot (\frac{π}{2}) =$ $tan 10x + cot 10x = tan (pi/2) + cot (pi/2) =$ Inf. + 0 = Infinity (rejected)
b. $tan x = \frac{1}{2}$ $tan x = 1/2$ --> $x = 26^{\circ} 565$ $x = 26^@565$
$10 x = 265^{\circ} 65$ $10x = 265^@65$
$tan 10 x + cot 10 x = tan (265.65) + cot (265.65) =$ $tan 10x + cot 10x = tan (265.65) + cot (265.65) =$
$= 13.15 + 0.08 = 13.23$ $= 13.15 + 0.08 = 13.23$

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Question #18b64

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