21X=tan(p)
Y=tan(q)
LHS = tan(p+q);
p+q = tan-13
tan-1x + tan-1y = tan-13;
u can take it frm then on.....
question edited::
find the no. of positive integral pairs of the solution of (x+y)/(1-xy) = 3
i know this shud be in algebra but actually this is one step of an inverse trig. question....
options:
(a) one (b) zero (c) two (d) none of these
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10 Answers
21
106the actual question was :
the number of positive integral solutions of the equation
tan-1x + cos-1y/√1+y2 = sin-13/√10
I simplified to tan-1x + tan-1y = tan-13
i.e. (x+y)/(1-xy) = 3
but i think positive integral solutions is zero....
i dont know answer this was a question in Brilliant's test 2day.... thx sky....
1well...
i dunno wat u did then...
do it simply naa...
separate y and x terms:
u will get y= 3-x/1+3x
now, see for which x, u get integral y's...
there wont be many as u can c, bcoz x>0 or <-2 will give a fraction..
