9asish is rite
q.73 if \frac{tan3x-tan2x}{1+tan3xtan2x}=1 0<x<pi/2 then x is
a)pi/3
b)pi/4
c)pi/2
d)solution does not exist
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18 Answers
1in d expression \frac{tan\frac{3\prod{}}{4} - tan\frac{\prod{}}{2}}{1+tan\frac{3\prod{}}{4}tan\frac{\prod{}}{2}}
dividing numerator and denominator with tanÎ /2 will give the value of tanÎ /4 as 1...
so amit is correct rite !! [12]
106@amit: tan(a-b) = (tana-tanb)/(1+tanatanb) is defined only when tan(a-b), tana and tanb are defined..
@race: tanpi/2 is NOT DEFINED.. rather tanpi/2- = ∞ and tanpi/2+ = -∞ where pi/2+ means just greater than pi/2 and pi/2- means just less than pi/2
1tanÎ /2 is a number tending to a very large value (we say, infinity) so there's nothing wrong if we divide it in numerator n denominator...
106@amit and race: What is tan(Î /2) ?? is it defined ??? ur all using tanpi/2 but check whether it is defined
1\frac{tan3\Pi /4-tan\Pi /2}{1+tan3\Pi /4.tan\Pi /2}
the above expression is of the form tan(a-b)= tana -tanb/1+tana.tanb
which will give tan(3pi/4-pi/2)=tan(pi/4) =1
106tan(3x-2x) = (tan3x - tan2x)/(1+tan3xtan2x) = 1
==> tanx=1
==> x=π/4 but 2x = π/2 hence tan2x is not defined hence solution does not exist
62see amit the thing is that at pi/4, the given function itself does not satisfy..
So the function is undefined..
1i have told to take x=pi/4 (edited now)
which satifies the equation
62amit.. does asish solution not look correct?!!
I did not get what your proof is..
put x=1 it does not satisfy
1take x=pi/4\frac{tan3\Pi /4-tan\Pi /2}{1+tan3\Pi /4.tan\Pi /2}
then using tan(a-b) formula
above expression will become tan(3pi/4 - pi/2)= tan(pi/4)=1
hence x=pi/4