Solve Sec^2x - 1 = 1/cot(x)? Interval of x is [0, 360)

Question

${sec}^{2}$ $sec^2$ x - 1 = 1 / cot(x)

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Answer 1 · 2018-05-28T04:15:01

$x = 0 or 90$ $x = 0 or 90$

First, we use Pythagorean identities.

${sec}^{2} (x) - 1 = {tan}^{2} (x)$ $sec^2(x) - 1 = tan^2(x)$

${tan}^{2} (x) = tan (x)$ $tan^2(x) = tan(x)$

We now have a polynomial in $tan (x)$ $tan(x)$ .

${tan}^{2} (x) - tan (x) = 0$ $tan^2(x) - tan(x) = 0$

$tan (x) (tan (x) - 1) = 0$ $tan(x)(tan(x)-1) = 0$

So, $tan (x) = 0$ $tan(x) = 0$ or $tan (x) = 1$ $tan(x) = 1$ .

$x = 0 or 90$ $x = 0 or 90$ .