# Thread: trig help / simplifying

1. ## trig help / simplifying

Given that $tan^{2}K = 2secK$

a) find the value of sec k.
(b) deduce that cosk=√2−1

$tan^{2}k = 2secK$

$sec^2k-1 = 2secK$

$sec^2K-2secK-1 = 0$

$secK = 1+ \sqrt{2}$

b) $\frac{1}{cosk} = 1+\sqrt{2}$

$cosK = \frac{1}{1 + \sqrt{2}}$

I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by $1 - \sqrt{2}$ I still don't get $cosK = \sqrt{2} -1$

2. Originally Posted by Tweety
Given that $tan^{2}K = 2secK$

a) find the value of sec k.
(b) deduce that cosk=√2−1

$tan^{2}k = 2secK$

$sec^2k-1 = 2secK$

$sec^2K-2secK-1 = 0$

$secK = 1+ \sqrt{2}$

b) $\frac{1}{cosk} = 1+\sqrt{2}$

$cosK = \frac{1}{1 + \sqrt{2}}$

I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by $1 - \sqrt{2}$ I still don't get $cosK = \sqrt{2} -1$
$\frac{1}{1+\sqrt{2}}\cdot\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{1-\sqrt{2}}{1-(\sqrt{2})^2}=\frac{1-\sqrt{2}}{1-2}=\frac{1-\sqrt{2}}{-1}=\sqrt{2}-1$

$
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$tan^{2}k = 2secK
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$
$
$sec^2k-1 = 2secK
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$sec^2K-2secK-1 = 0
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after that if we take $\sec k = x$
then
$x^2-2x-1=0$
$x=1$
$\sec x=1$

4. jashansinghal: your mistake is having the color tags inside of the math tags in LaTex...

Oh, you mean the problem?
$x^2-2x-1=0$
$x=1$
$\sec x=1$
I don't know why you got x = 1 from the quadratic. You need to use the quadratic formula.
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 4}}{2}$
$x = \frac{2 \pm 2\sqrt{2}}{2}$
$x = 1 \pm \sqrt{2}$

Reject $x = 1 - \sqrt{2}$ because secent of an angle cannot be in between -1 and 1.
$x = 1 + \sqrt{2}$
$\sec k = 1 + \sqrt{2}$

01

5. okkk sorry