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If where xy < 0, then the possible values of t is/are
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a
4π5
b
7π10
c
9π10
d
21π20
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Correct option is C
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−1x+1x∈π3,π2,sec−1y+1y∈π2,2π3⇒sec−1x+1x˙+sec−1y+1y∈5π6,7π6Similarly for second case also, t∈5π6,7π6
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