(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

(1+tan1)(1+tan2)(1+tan3) .................(1+tan45) = 2ⁿ

                                                                                        (1+tan45 = 1+1=2)

(1+tan1)(1+tan2) ........................(1+tan44) = 2^n-1

now make pairing such that sum of angles is 45...

[(1+tan1)(1+tan44)][(1+tan2)(1+tan43) ...................[(1+tan22)(1+tan23)] = 2^n-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 = 2^n-1

k²² = 2^n-1

2²² = 2^n-1

n = 23 ans

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Approved