'104 Mr. Adams on I [Feb. 



i«i x= tan. « = ^ = + /x = A +/« .*. A = 0, 

 t^:=.\ + r = l+J\z^B+f,z .-.B = 1, 



^ =2^ + 2 ^^=0+/„;« = 2C +/.;.. -.0=0, 



^ = 2 4- 8/« + 6i*= 2 4-:/,T« = 2.3D+/,.,«.-.D 



"" 1.2.3' 



^ = 16^ + 40^^ -r 24^^ =0+y> = 2.3 .4E4-y>.-. 



E = 0, 

 ^ = 16 + 136 r- + 240 ^^ + 120 ^« = 16 +y; ;« = 2 . 3 . 



4.5F+/,....F = ^£^, 



&c 



From whence, and by substituting in the problem, we have 



*"»• ^ = ^ + r^ + 170^475 + ^"=- 



Example 13. — To find the secant of an arc in terms of the arc, 

 radius unity. - C . 

 u — sec. z — I A- fz = A +/;?;.•. A = 1, 



3; .-. D = 0, 



;^ = 5+-g-+-^ + &c. = 5+/,;2 = 2.3.4 £«+ /; 



"^ •*• ^ = irlri' 



'^* " ^1 ^ ■ 24U3 . 641 <5 , Q A . r o o /« c 



^-j^^ = 6 W + -g- 4- -^ ± &c. = + ^5 ^ = 2 . 3 . 4. 5 



F+/,^.-.F=0, 

 £^=61 + ?^± &c.= 61 +/e 25 = 2.3.4.5.6 0' + 



&c 



• •...•... «-»•«« •'»'•• »» >•♦ ••• •••». 



t" 6 8* 61 



Hence we have sec. « = 1 + r— ^ + 



1.2 '1 .8.3.4 ' 1 .2.3.4.5.5 

 + &C. 



