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

## Comments

Assignment Expert08.11.13, 10:06There'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.

curious03.11.13, 10:25(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?

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