If a1, a2,  ..., an  are in A.P with common difference d≠0,  then (sin  d)  [sec  a1 sec  a2+sec  a2 sec  a3+ ... +sec  an−1sec  an] is equal to

Given,  common difference =d=a2−a1=a3−a2=a4−a3= .......... =an−an−1 ∴ (sin  d)  [sec  a1 sec  a2+sec  a2 sec  a3+ ... +sec  an−1sec  an]=sindcos  a1 cos a2+sindcos  a2 cos a3+  ...  +sindcos  an−1 cos an=sin(a2−a1)cos  a1 cos a2+sin(a3−a2)cos  a2 cos a3+  ...  +sin(an−an−1)cos  an−1 cos an     since   sina2-a1cos a1 cosa2= sin a2 cosa1- cosa2sina1cos a1cos a2 = tan a2-tan a1  , similarly the remaining terms =tan a2−tan a1+tan a3−tan a2+ ........ +tan an−tan an−1=tan an−tan a1