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Question

Given that (1+tan1)(1+tan2)....(1+tan45)=2, find n.

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Solution

Given that:(1+tan1)(1+tan2)........(1+tan45)=2n
LHS,
(1+tan1)(1+tan2)........(1+tan45)(A)
we know that
tan(A+B)=tanA+tanB1tanA.tanB, also, tan45=1
So,
tan45=tan1+tan441tan1.tan44tan1+tan44(1tan1.tan44)(tan45)(i)tan45=tan2+tan431tan2.tan43tan2+tan43(1tan2.tan43)(tan45)(ii)....tan45=tan22+tan231tan22.tan23tan22+tan23(1tan22.tan23)(tan45)(22)
Using the above equation we can simplify the expression (A)
(1+tan1)(1+tan2)(1+tan3)........(1+tan44)(1+tan45)=(1+tan1)(1+tan2)(1+tan3)........(1+tan44)×2=2(1+tan1)(1+tan2)(1+tan3)........(1+tan44)=2(1+tan1)(1+tan44)(1+tan2)(1+tan43)(1+tan3)(1+tan42).....(1+tan22)(1+tan23)=2(tan1+tan44+tan1.tan44+1)(tan2+tan43+tan2.tan43+1).......(tan22+tan23+tan22.tan23+1)=2[(tan45)(1tan1.tan44+tan1.tan44+1)(tan45(1tan2.tan43)+tan2.tan43+1).......(tan45(1tan22.tan23)+tan22.tan23+1)]=2[(1tan1.tan44+tan1.tan44+1)(1tan2.tan43+tan2.tan43+1).......(1tan22.tan23+tan22.tan23+1)]=2.(2).(2)..........(2)=2.222=223n=23

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