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

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

Grade:

2 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 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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Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.

All the best Akash!!!

Regards,
Askiitians Experts
Priyansh Bajaj

Kushagra Madhukar
askIITians Faculty 629 Points
9 months 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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