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ADNAN MUHAMMED

Grade 12,

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(1+TAN1)(1+TAN2).....................(1+TAN45) =?

(1+TAN1)(1+TAN2).....................(1+TAN45)
=?

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2 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
13 years ago

Dear Akash,

Solution:-  Club the terms like this-

[(1+tan1)(1+tan44)] * [(1+tan2)(1+tan43)] * [(1+tan3)(1+tan42)].......[(1+tan22)(1+tan23)]*[(1+tan45)]

= [(1+(tan1+tan44) + (tan1)(tan44)] * [(1+(tan2+tan43) + (tan2)(tan43)]......[(1+(tan22+tan23) + (tan22)(tan23)]* 2

Since, tanA + tan B = tan(A+B) [1- (tanA)(tanB)]. So, (tan1 +tan44) = tan(45) [1- (tan1)(tan44)] = [1- (tan1)(tan44)]. Doing it for all terms and putting them in above expression will give-

[(1+(1- (tan1)(tan44)) + (tan1)(tan44)] * [(1+(1- (tan2)(tan43)) + (tan2)(tan43)]......[(1+(1- (tan22)(tan23)) + (tan22)(tan23)]* 2

= (2)*(2)*(2)......(2)*(2) = 223 [ANS].                        

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Regards,
Askiitians Experts
Priyansh Bajaj

Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
[(1+tan1)(1+tan44)] * [(1+tan2)(1+tan43)] * [(1+tan3)(1+tan42)].......[(1+tan22)(1+tan23)]*[(1+tan45)]
= [(1+(tan1+tan44) + (tan1)(tan44)] * [(1+(tan2+tan43) + (tan2)(tan43)]......[(1+(tan22+tan23) + (tan22)(tan23)]* 2
 
Since, tanA + tan B = tan(A+B) [1- (tanA)(tanB)]. So, (tan1 +tan44) = tan(45) [1- (tan1)(tan44)] = [1- (tan1)(tan44)]. Doing it for all terms and putting them in above expression will give-
[(1+(1- (tan1)(tan44)) + (tan1)(tan44)] * [(1+(1- (tan2)(tan43)) + (tan2)(tan43)]......[(1+(1- (tan22)(tan23)) + (tan22)(tan23)]* 2
= (2)*(2)*(2)......(2)*(2) = 223
 
Thanks and regards,
Kushagra

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