Prove that $t a n 18^{\circ}$ is a root of the equation $5 x^{4} - 10 x^{2} + 1 = 0$ and hence prove that $t a n^{2} 18^{\circ}$ 4 = 0.1056

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Solution

If $θ = 18^{\circ} 5 θ = 90^{\circ}$ $3 θ = 90^{\circ} - 2 θ$
or $t a n 3 θ = c o t 2 θ o r \frac{3 t - t^{3}}{1 - 3 t^{2}} = \frac{1 - t^{2}}{2 t}$
$∴ 5 t^{4} - 10 t^{2} + 1 = 0$ \, \, \, $∴ t^{2} = 1 \pm \sqrt{0.8}$
$o r t^{2} = 1 \pm .8944 = 1 - .8944 = 0.1056$
$t^{2} = t a n^{2} 18^{\circ}$ is +ive and less than 1.
$t a n^{\circ} 45 = 1 ∴ t a n 18^{\circ} < t a n 45^{\circ} o r t < 1$ and +ive . The other value being > 1 is rejected

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