The correct option is C (2nπ−π2,2nπ+π2)−{2nπ},n∈Z ; (0,π2)
Since in cosec−1 t
t∈(−∞,−1]∪[1,∞)
So, √t∈[1,∞)
√log(3−4secx1−2secx)2 ≥1
Squaring both sides,
log(3−4secx1−2secx)2≥1
and base of log
1<3−4secx1−2secx<2
Now, 3−4secx1−2secx<2
⇒3−4secx1−2secx−2<0
⇒11−2secx<0
⇒1−2secx<0
⇒secx>12 ⋯(1)
(If secx=1, then base is 1 ⇒x≠ 2nπ)
And
3−4secx1−2secx>1
⇒3−4secx1−2secx−1>0
⇒1−secx1−2secx>0
⇒secx−12secx−1>0
⇒secx−1>0
⇒secx>1 ⋯(2)
Clearly, from equation (1) and (2),
x must lie in 1st or 4th quadrant.
So, domain =(2nπ−π2,2nπ+π2)−{2nπ}
and the range is (0,π2) since t does not take negative values.