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

Krystal jain , 7 Years ago
Grade 12
anser 2 Answers
Dhruvit Raithatha

Last Activity: 7 Years ago

\\(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

Rishi Sharma

Last Activity: 4 Years ago

Hello students,
The solution of the above problem is attached.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.645-357_WhatsApp Image 2020-06-06 at 1.32.02 PM(1).jpeg

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