Question

If a1, a2, a3, ..., an, are in A.P. whose common difference is d, show thatsec a1 sec a2 + sec a2 sec a3 + ... + sec an . sec a1 = (tan an- tan a1)/sin d​

Answer 1

sec a1 sec a2 + sec a2 sec a3 + ... + sec an-1 . sec an = (tan an- tan a1)/sin d​  if If a1, a2, a3, ..., an, are in A.P with d common difference

Step-by-step explanation:

If a1, a2, a3, ..., an, are in A.P.

common difference is d,

sec a1 sec a2 + sec a2 sec a3 + ... + sec an-1 . sec an = (tan an- tan a1)/sin d​

=> Sind (sec a1 sec a2 + sec a2 sec a3 + ... + sec an-1 . sec an) =  (tan an- tan a1)

LHS

= Sind (sec a1 sec a2 + sec a2 sec a3 + ... + sec an-1 . sec an)

= Sind * sec a1 sec a2   + Sind*sec a2 sec a3 + ............................+ Sind*sec an-1 . sec an

d = a2 - a1  = a3 - a2  = an - an-1  

= Sin(a2 - a1) *(sec a1 sec a2)   + Sin(a3 - a2)*sec a2 sec a3  +..............................+Sin(an - an-1)*sec an-1 . sec an

= (Sina2Cosa1 - Cosa2Sina1)* sec a1 sec a2  + (Sina3Coa2 - Cosa3Sina2)* sec a2 sec a3  +.............................+ (SinanCoan-1 - CosanSinan-1)/ sec an-1 . secan

cosa * Seca = 1   & sina * Seca = Tana

=  (Tan a2 - Tana1)  + (tana3 - Tana2)  +...................................+(Tanan - Tanan-1)

= Tanan  -  Tana1

= RHS

= QED

Proved

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