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

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

Grade:12

2 Answers

Dhruvit Raithatha
32 Points
5 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
askIITians Faculty 646 Points
2 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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