If `(tan^-1x)^2+(cot^-1x)^2=(5pi^2)/8`, then x equals

Answer

`x = -1`

Explanation

`(tan^(-1) x)^(2) + (cot^(-1) x)^(2) = (5pi^(2))/(8)`
`rArr (tan^(-1)x + cot^(-1) x)^(2) 2 tan^(-1) x ((pi)/(2) - tan^(-1) x) = (5pi^(2))/(8)`
`rArr (pi^(2))/(4) - 2 xx (pi)/(2) tan^(-1) x + 2 (tan^(-1) x)^(2) = (5pi^(2))/(8)`
`rArr 2(tan^(-1) x)^(2) - pi tan^(-1) x - (3pi^(2))/(8) = 0`
`rArr tan^(-1) x = -(pi)/(4), (3pi)/(4)`
`rArr tan^(-1) = -(pi)/(4) " " ( :' tan^(-1) x in (-(pi)/(2), (pi)/(2)))`
`:. x = -1`