The correct option is D {−3π2,5π2}
f(x)=4tan−1x+3sin−1x+sec−1x
Domain of tan−1x is R …(1)
Domain of sin−1x is [−1,1] …(2)
Domain of sec−1x is (−∞,−1]∪[1,∞) …(3)
Hence, domain of f(x) is (1)∩(2)∩(3)
⇒x={−1,1}
Now, f(−1)=4tan−1(−1)+3sin−1(−1)+sec−1(−1)
=4(−π4)+3(−π2)+π=−3π2
f(1)=4tan−1(1)+3sin−1(1)+sec−1(1)
=4(π4)+3(π2)+0=5π2
Hence, range of f(x) is {−3π2,5π2}.