Question

it tan²x + cot²x = 2, find x where 0⁰ < x⁰ < 90⁰​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ value \: of \: x \: (inclination \: ) \\ \\ given \: that : \\ tan {}^{2} \: x + cot {}^{2} \: x = 2 \\ \\ tan {}^{2} x + \frac{1}{tan {}^{2} \: x } = 2 \\ \\ let \: here \: then \\ tan {}^{2} \: x = p \\ \\ thus \: then \\ \: resubstituting \: \: accordingly \\ \: we \: get \\ p + \frac{1}{p} = 2 \\ \\ \: p {}^{2} + 1 = 2p \\ ∴ \: p {}^{2} - 2p + 1 = 0 \\∴ \: p {}^{2} - p - p + 1 = 0 \\ \\ p(p - 1) - 1(p - 1) = 0 \\ (p - 1)(p - 1) = 0 \\ \\ p - 1 = 0 \\ ∴ \: p = 1$

$thus \: then \\ tan {}^{2} \: x = p \\ tan {}^{2} \: x = 1 \\ \\ tan \: x = 1 \: \: or \: \: tan \: x = - 1 \\ but \: if \: we \: take \\ tan \: x = - 1 \\ then \: we \: have \\ x \: would \: lie \: in \: either \: in \\ second \: or \: fourth \: quadrant \\ and \: since \: here \\ 0 < x < 90 \\ \\ tan \: x = - 1 \: \: is \: absurd \\ hence \: then \\ tan \: x = 1 \\ \\ we \: know \: that \\ tan \: 45 = tan \: \frac{\pi}{4} = 1 \\ \\ ∴ \: \: x = \frac{\pi {}^{c} }{4}$