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```        (1+tan 1)(1+tan 2)(1+tan 3)+(1+tan 4)..................(1+tan 45)=2^n(two to the power n)
find the value of n```
6 years ago

510 Points
```										(1+tan1)(1+tan2)(1+tan3) .................(1+tan45) = 2n
(1+tan45 = 1+1=2)
(1+tan1)(1+tan2) ........................(1+tan44) = 2n-1
now make pairing such that sum of angles is 45...
[(1+tan1)(1+tan44)][(1+tan2)(1+tan43) ...................[(1+tan22)(1+tan23)] = 2n-1

now take any of pair separately ,let its value be k then
k=(1+tan1)(1+tan44) = 1 + tan44 + tan1 + tan1tan44         .................1
tan45 = tan(44+1) = (tan44+tan1)/(1-tan1tan44)          .......................2
from eq2 put  ( tan1+tan44) in eq 1
k=1 + tan45 = 2
now there will be 22 pairs  so the value of expression is
k.k.k.k............22terms = 2n-1
k22 = 2n-1
222 = 2n-1
n = 23 ans
approve my ans if u like

```
6 years ago
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