"ィ=Em( ) = Eo1ere, Elm'25+D-m @-3 1+47-1 r=tan-'3_tan-']+tan-'5-tan-"3+.....+ tan"(2n+1)-tan"'(2n-1)+""""+o t=lim|tan "2n+1)-tan']= lim tan -I 2n lim tar TTP a [1+(2n+1) t=tan'(= # tant=1 (i)=(a) (i) . a、b,c are in AP = 2b=a+c cos0 =- a 1-tan`0,/2 a tan2 b+c-a b+c - 1+ian'8,/2 btc 2 htc4a Similarly tan20_a+b-c +tan?0 2r tanon 2b 2 ⁃ 1 (ii)=(c) 2 a+b+c 2 2 a+b+c 3b 3 (ii) Equation of line through (0,1,0) & perpendicular to x+2y+22-0 is X-2-1-E-A 1 2 2 (1,21+1,22) For some value of d,the foot of perpendicular from origin to line $ 2 2.121+1+2-21=0 2 1=- ." Foot of perpendicular is XA .". Required distance= 425 16 /45 5 + +- I V818188i 3 (iii)= Solve this and you're a genius