An algebra problem by Siddhant Mahurkar

Algebra Level 1

The value of tan 1 ° × tan 2 ° × tan 3 ° × × tan 89 ° \tan 1° \times \tan 2° \times \tan 3°\times \ldots \times \tan 89° is

0 1 -1 1/2

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3 solutions

Mayank Holmes
May 26, 2014

tan1 tan2 tan3 ...............tan44 = cot46 cot47 cot48.........cot89 = ( 1/ (tan46 *tan47 *tan48 ...... ................................... tan89 )............. therefore this whole sequence is equal to nothing but tan(45) ........which is equal to 1 !!!!!!!!

Shathi Cse
Apr 18, 2014

tan1.tan2.........tan89 here, tan 89 can be written as cot 1[tan(90-1)=cot 1] by tan(90-theta)=cot(theta) likewise......, tan 88 =cot 2 tan 87= cot 3 ................up to tan 46=cot 44 then middle one is tan 45 =1 so it is tan1.tan2...........tan44.tan 45.cot44..............cot1 tan and cot cancel out by[tan (theta)*cot(theta)=1] so ....remaining is 1 answer =1

Satish Vootukuru
Apr 16, 2014

tan(90-theta)=cot(theta); tan1 * tan 89 = tan1 * tan(90-1) = tan1 * cot1 = tan1 (1/tan1) = 1; similarly tan2 tan88 = 1,tan3 tan87=1.....; 1 1 1.... tan45 = 1 since tan45 = 1

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