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If(tan 46°-tan1°-1)(tan 47°-tan2°-1)(tan 48°-tan 3°-1).........(tan 89°-tan 44°-1)=2 n .,then find the value of n.

  • If(tan 46°-tan1°-1)(tan 47°-tan2°-1)(tan 48°-tan 3°-1).........(tan 89°-tan 44°-1)=2n.,then find the value of n.

Grade:12

2 Answers

Arun
25750 Points
5 years ago
Dear Kaushal
 
tan(45 + 1) – tan 1 – 1
 
 = 1 + tan1 / (1 – tan1) – tan 1 – 1
 
 = (tan^2 1 + tan1) = tan1 (1 + tan1)
 
similarily other terms will become
 
(tan 1* tan2 * tan3 *......) (1 + tan1)(1 + tan2)(1 + tan3.......)(1 + tan43)(1+ tan44)
 
 = 1 * 2*2*2*2* …...22 times
 
2^22
 
n = 22
ketav
13 Points
3 years ago
tan(46-1)=(tan 46- tan 1)/1+tan 1.tan46
1+tan1.tan46=tan46-tan1
tan46-tan1-1=tan1.tan46
similarly,
tan47-tan2-1=tan2.tan47
thus, the series becomes
tan1.tan2.tan3.....................tan89
(tan89=cot1)
=tan1.cot1.tan2.cot2..........................tan44.cot44.tan45
=1.tan45
=1=2n
20=1
n=0
 
 

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