Question

If (tan⁻¹x)² + (cot⁻¹x)² = 5π²/8, then find x.

Answer 1

 

(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`