InterviewSolution
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InterviewSolution
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If `a_1,a_2,a_3, ,a_n`are in A.P. withcommon difference `d(w h e r ed!=0)`, then the sum ofseries.`s in dd(cos e ca_1cos e ca_2+cos e ca_2cos e ca_3+=cos e ca_(n-1)cos e ca_n)`is equal to`cota_1-cota_ndot` |
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Answer» Since ` a_(1) , a_(2),a_(3) ,….,a_(n)` are in AP with common difference d, we have ` (a_(2)-a_(1)) = (a_(3)-a_(2)) = (a_(4)-a_(3))= ….= (a_(n) - a_(n-1)) =d ` Sin d ` ( cosec a_(1) coses a_(2) + cosec a_(2) cosec a_(3) +……+ cosec a_(n-1) cosec a_(n)` ) ` sind/(sina_(1)sina_(2))+ sind/(sina_(2) sina_(3))+sind/(sina_(3)sina_(4))+......+sindd/(sina_(n-1)sina_(n))` ` sin (a_(2) -a_(1))/(sina_(1) sin a_(2)) + sin (a_(3) -a_(2))/ (sin a_(2) sina_(3)) + sin( a_(4) -a_(3))/(sin a_(3) sin a_(4))/+ ....+ sin(a_(n)-a_(n-1))/(sin a_(n-1) sin a_(n)) ` ` [ therefore d= (a_(2) -a_(1) = ( a_(3) -a_(2)) =...= (a_(n) -a_(n -1)]` ` ( sina_(2) cos a_() - cos a_(2) sin a_(1))/(sin a_(1) sin a_(2)) + (sin a_(3) cos a_(2) -cos a_(3) sin a_(2))/ (sin a_(2) sin a_(3)) + ( sin a_(n) cos a_(n-1) -cos a_(n) sin a_(n-1))/ (sin a_(n-1) sina_(n))` ` = ((cos a_(1))/ (sin a_(1)) - (cos a_(2))/sin a_(2)) + ((cos a_(2))/(sin a_(2)) - (cos a_(3))/ (sin a_(3))) +......+ ((cos a_(n-1))/(sin a_(n-1)) - (cos a_(n))/(sin a_(n)))` ` ( cot a_(1) - cot a_(2)) + ( cot a_(2) -cot a_(3)) + (cot a_(3) - cot a_(4)) + .....+ ( cot a_(n-1) -cot a_(3)) + (cot a_(3) - cot a_(4)) + ....+ (cot a_(n-1) - cota_(n))` ` = ( cot a_(1) -cot a_(n))` Hence, sin d ` (coses a_(1) cosec a_(2) + cosec a_(2) cosec a_(3) + .....+ cosec a_(n-1) cosec a_(n)) = ( cot a_(1) -cot a_(n))` |
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