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Logarithmic Differentiation

If tan-1 4 ...

Question

If $t a n^{- 1} 4 = 4 t a n^{- 1} x$ , then $x^{5} - 7 x^{3} + 5 x^{2} + 2 x$ is equal to

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Solution

$4 = t a n 4 θ$ where $x = t a n θ$
$t a n 2 θ = \frac{2 x}{1 - x^{2}} \Rightarrow t a n 4 θ = \frac{2 (\frac{2 x}{1 - x^{2}})}{1 - \frac{4 x^{2}}{(1 - x^{2})^{2}}}$
$= \frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}} = 4$ (given)
$\Rightarrow x^{4} + x^{3} - 6 x^{2} - x + 1 = 0$
Now, $x^{5} - 7 x^{3} + 5 x^{2} + 2 x = (x - 1) (x^{4} + x^{3} - 6 x^{2} - x + 1) + 1 = 1$

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