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Question
If (tan−1x)2 + (cot−1x)2 = 5π2/8, then find x.
Solution
(tan−1x)2 + (cot−1x)2 = 5π2/8
`=>(tan^(−1)x+cos^(−1)x)^2−2tan^(−1)xcot^(−1)x=(5π^2)/8`
`⇒(π/2)^2−2tan^(−1)x(π/2−tan^(−1)x)=(5π^2)/8`
`⇒π^2/4−πtan^(−1)x+2(tan^(−1)x)^2=(5π^2)/8`
`⇒2(tan^(−1)x)^2−πtan^(−1)x+π^2/4−(5π^2)/8=0`
`⇒2(tan^(−1)x)^2−πtan^(−1)x−(5π^2+2π^2)/8=0`
`⇒2(tan^(−1)x)2−πtan^(−1)x−(3π^2)/8=0`
Solving the quadratic equation, we get
`⇒tan^(−1)x=(π±sqrt(π^2+4xx2xx(3π^2)/8))/(2xx2)`
`⇒tan^(−1)x=(π±2π)/4`
`⇒tan^(−1)x=(3π)/4 or tan^(−1)x=−π/4 `
`⇒x=tan((3π)/4) or x=tan(−π/4)`
`⇒x=−1`
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