Question

plzzz ans this ques

Answer 1

[(1+tan1°)(1+tan44°)] [(1+tan2°)(1+tan43°)] ...............up to 22 pairs × (1+tan45°) =2^n,



[1+tan44°+tan1°+tan1°.tan44°] [1+tan2°+tan44°+tan2°.tan44°]................up to 22 pairs ×(1+1) =2^n, ................eq(1),




since
in each of the brackets sum of both the angles are 45°,

then
we can say that
tan(1°+44°)=(tan1°+tan44°)/(1-tan1°.tan44°),

then

1 =(tan1°+tan44°)/(1-tan1°.tan44°),

then

tan1°+tan44°=1-tan1°.tan44°,

similarly for other brackets,

so from eq(1), we have


[1+1-tan1°.tan44°+tan1°.tan44°] [1+1-tan2°.tan43°+tan2°.tan43°].........upto 22 pairs × 2 =2^n,

[1+1] [1+1]........upto 22 pairs × 2 =2^n,



then

2×2×.......upto 22 pairs × 2= 2^n,


2²² × 2 =2^n,


therefore

2²³=2^n,


then

n=23