Arun8 Years agoDear student dy/dx = [1 /1 + (secx + tanx)² ][secx tanx + sec²x] dy/ dx = sec x(secx + tanx) /(1+ sec² x + tan²x + 2secx tanx) dy/dx = sec x(secx + tanx) / 2secx ( secx + tanx) dy/dx = ½ Hence d²y /dx² = 0
Sounak Dutta8 Years agoY=tan-¹(secx+tanx)Y=tan-¹((1+sinx)/cos)Y=tan-¹((sinx/2+cosx/2)/(cosx/2-sinx/2))Y=tan-¹(tan(pi/4+x/2)Y=pi/4+x/2dy/dx=1/2d²y/dx²=0