The correct option is D 2
(1−tanA+secA)(1−cotA+cosecA)
= (1−sinAcosA+1cosA)(1+cosAsinA+1sinA)]
= (cosA−sinA+1)(sinA−cosA+1)sinAcosA
= cosAsinA−cos2A+cosA−sin2A+sinAcosA−sinA+sinA−cosA+1sinAcosA
= 2cosAsinA+1−(sin2A+cos2A)sinAcosA
= 2 (Since, sin2A+cos2A=1)