If t=sec−1⁡x+1x+sec−1⁡y+1y where xy < 0, then the possible values of t is/are

Since xy < 0 ⇒x <0, y> 0 or x> 0, y <0x+1x≥2,y+1y≤−2 (or) x+1x≤−2,y+1y≥2according to the case for first case sec−1⁡x+1x∈π3,π2,sec−1⁡y+1y∈π2,2π3⇒sec−1⁡x+1x˙+sec−1⁡y+1y∈5π6,7π6Similarly for second case also, t∈5π6,7π6