Question #17377
(1+tan1)(1+tan2)................(1+tan45)?
Expert's answer
if a+b=45 then (1+tan(a) )( 1+ tan(b) )=2
tan(b)=tan(45-a) =2 cos(a)/(cos(a)+sin(a))-1
1+tan(b)=2 cos(a)/(cos(a)+sin(a))
tan(a)=sin(a)/cos(a)
1+tan(a)=1+sin(a)/cos(a)=(sin(a)+cos(a))/cos(a)
(1+tan(a) )( 1+tan(b) )=[ (sin(a)+cos(a))/cos(a) ]*[2cos(a)/(cos(a)+sin(a))]=2
(1+tan1)(1+tan2)................(1+tan45)=[(1+tan1)(1+tan44) ]*[ (1+tan 2)(1+tan43) ]*...*[(1+tan22)(1+tan23)] *(1+tan
45)=2^(22) * (1+tan 45)=2^23
tan(b)=tan(45-a) =2 cos(a)/(cos(a)+sin(a))-1
1+tan(b)=2 cos(a)/(cos(a)+sin(a))
tan(a)=sin(a)/cos(a)
1+tan(a)=1+sin(a)/cos(a)=(sin(a)+cos(a))/cos(a)
(1+tan(a) )( 1+tan(b) )=[ (sin(a)+cos(a))/cos(a) ]*[2cos(a)/(cos(a)+sin(a))]=2
(1+tan1)(1+tan2)................(1+tan45)=[(1+tan1)(1+tan44) ]*[ (1+tan 2)(1+tan43) ]*...*[(1+tan22)(1+tan23)] *(1+tan
45)=2^(22) * (1+tan 45)=2^23
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#339914 on Dec 2023
#339914 on Dec 2023
Comments
There's no b in the RHS because we substitute tan(b)=tan(45-a) =2 cos(a)/(cos(a)+sin(a))-1 into the expression instead of tan b. The main idea of the solution is that we can gather pairs of angles which will give 45 summed up, like say 1+44, 2+43 and so on.
(1+tan(a) )( 1+ tan(b) )=[ (sin(a)+cos(a))/cos(a) ]*[2cos(a)/(cos(a)+sin(a))]=2 how did this happen... why is there no 'b' in the RHS ?? can u please give the detailed steps of the above equation?