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乖(1 + tan 1°)(1 + tan 2)(1 + tan 45°) = 2", |
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Answer» =(1+tan 0o)(1 + tan1°) (1 + tan2°) (1 + tan3°) ………. (1 + tan 45°) {Since , tan 0o= 0} Now , we know thattan(A + B) = [ tan(A) + tan(B) ] / [ 1 - tan(A) tan(B) ]And tan(45) = 1 Sotan(45) = [ tan1 + tan44 ] / [ 1 - tan1 tan44 ]tan(45) = [ tan2 + tan43 ] / [ 1 - tan2 tan43 ]...tan(45) = [ tan22 + tan23 ] / [ 1 - tan22 tan23 ] So we have ,tan1 + tan44 = tan(45) [ 1 - tan1 tan44 ]tan2 + tan43 = tan(45) [ 1 - tan2 tan43 ]...tan22 + tan23 = tan(45) [ 1 - tan22 tan23 ] So finally we have , (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + tan45) = (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + 1) = 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)(1 - tan1 tan44) + tan1 tan44 + 1) ((tan45)(1 - tan2 tan43) + tan2 tan43 + 1) ... ((tan45)(1 - tan22 tan23) + tan22 tan23 + 1) = 2 (1 - tan1 tan44 + tan1 tan44 + 1) (1 - tan2 tan43 + tan2 tan43 + 1) ... (1 - tan22 tan23 + tan22 tan23 + 1) = 2 (2) (2) ... (2) = 2 (2^22) = 2^23 hence n=23 |
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