If tan2x+secx−a=0 has alteast one solution, then a∈..........
(−∞,1]
[-1,1]
[1,∞)
(−∞,−1]
tan2x+secx−a=0(sec2x−1)+secx−a=0=sec2x+secx+14−1−a=0=(secx+12)=54+aNowsecx≥1or≤−1⇒secx+12≥32orsecx+12≤−12∴(Secx+12)2≥94or(Secx+12)2≥14i.e.54+a≥14=a≥−1
If tan2x+sec x−a=0 has alteast one solution, then a∈ ___
If esinx−e−sinx=a has alteast one real solution, then