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(tan inverse x )^2 + (cot inverse x ) ^2=5π^2/8.solve for x

Krystal jain , 8 Years ago
Grade 12
anser 1 Answers
Dhruvit Raithatha
\\(tan^{-1}x)^2 + \left( \frac{\pi}{2} - tan^{-1}x\right)^2 = \frac{5\pi^2}{8} \\\\ Let\ tan^{-1}x = k \\\\ 2k^2 -\pi k + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0 \\ 16k^2 -8\pi k - 3\pi^2 = 0 \\ 16k^2 -12\pi k+ 4\pi k - 3 \pi^2 = 0 \\ (4k-3\pi)(4k+\pi) = 0 \\ x = tan\left(\frac{3\pi}{4}\right), x = tan\left( \frac{-\pi}{4}\right) \\\\x = -1
Last Activity: 8 Years ago
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