Aspire Faculty ID #18389 · Topic: AMU MCA 2016 · Just now
AMU MCA 2016

A root of the equation $17x^2 + 17x \tan\left(2\tan^{-1}\frac{1}{5} - \frac{\pi}{4}\right) - 10 = 0$ is

Solution

Use identity: $\tan(2\tan^{-1}\frac{1}{5}) = \frac{2\cdot\frac{1}{5}}{1-\frac{1}{25}} = \frac{5}{12}$ Then $\tan(A - B) = \frac{\frac{5}{12}-1}{1+\frac{5}{12}} = -\frac{7}{17}$ Equation: $17x^2 -7x -10 = 0$ $(x-1)(17x+10)=0$ So $x=1$ Final Answer: $\boxed{1}$
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