21Sir,i was askin coz i got this un wrong : [2]
FT : p1 Q4
cot-1(-3)+cot-1(-2)=Ï€-cot-13+Ï€-cot-12=2Ï€-(cot-13+cot-12)=2Ï€-Ï€/4=7Ï€/4
wen appylying formulae lyk tan inv + tan invers = tan -1(x+y/1-xy)
can sum1 tell me if ther is a genreal rule for addtn/subtctn of Î wen negative n all dat........
OR IT HAS TO BE REMEMBERED CASE to CASE basiss??
-
UP 0 DOWN 0 0 14
14 Answers
62Dont remember case by case
instead always take tan of the whole thing on both sides...
if it is sin inverse then take sin of the whole thing
remembering the whole thing is alrite but i think not worrth spending that much time..
21
1x= (cot-13+cot-12)
x= tan-1(1/3)+cot-1(1/2)
tan x= tan(tan-1(1/3)+tan-1(1/2))
tan x= {tan(tan-1(1/3)+tantan-1(1/2))} / {1- tan(tan-1(1/3)tantan-1(1/2))}
tan x = (1/3+1/2)/(1-1/2.1/3) = 5/6 / (5/6) = 1
thus x = pi/4
21YE statement mera dbt hai bhaiyya not MY OWN working, to fir ye sab ans kyun???
cot-1(-3)+cot-1(-2)=Ï€-cot-13+Ï€-cot-12=2Ï€-(cot-13+cot-12)=2Ï€-Ï€/4=7Ï€/4
so the ans is 7pi/4
1cot-1(-3) + cot-1(-2)
= Î - cot-1(3) + Î - cot-1(2)
= 2Î - [cot-12+ cot-13 ]
= 2Î - cot-11
=2Î - Î /4 = 7Î /4
1:O :O :O :O :O :O
huh!!!
i din see only dat this is your sole doubt !!
i din read any of the above posts and started doing .... [actually kal hi main yeh chap thik se kii ... thanx to kamalendu :P ...]
1:) ... oh it was a great help ...
now i got insomnia! ... no sleep from yesday afternoon!
not tension ... but kal eto chap nie niechhilam naa ...
aar ami kichhu te chaap nie nile ... puro bhule jai sab kichhu :P ..
but i think i cud understand at the least 10% of the correct thing ... n
the whole credit goes to you .. :)
1CHOL TAHALE AJ RATE ABAR DEBO.....AMARO KOEKTA iit 2008 ER PAPER E DBT ACHCHE.......PARLE TYM WASTE NA HLE G TALK E ASHISH