1sin-1x goes from -pi/2 to pi/2, so does, tan-1x,
so put -pi/2 for min. x2+pix is increasing from -pi/2 to pi/2, so put -pi/2 for both tan-1y and sin-1x, to get k as 0
no ones sleeping, they are just busy with post result decision making :)
cheers!!
1357ajoy.. how did you go about trying Q 3?
So far how much have you tried..
This is more like the question where we need to solve for
1/m+1/n1-1/mn=1/λ
1I had got upto that ,How do we find the ordered pairs(m,n),by hit & trial method.?
1708tan-1(1/m) +tan-1(1/n) = tan-1(1/3)
tan-1{(m+n)/mn-1} = tan-1(1/3)
3m+3n = mn-1
or mn -1 =3m+3n
mn-3m -3n+9 - 10=0
m(n-3) - 3(n-3) = 10
(m-3)(n-3) =10
here m and n are Integer so (m-3) and (n-3) are also Integer.....
so product of 2 Integer is 10 which is possiable only when
(m-3)*(n-3) = 10*1 or 1*10
means (m-3)=10 and (n-3)=1
OR (m-3)=1 and (n-3) =10
OR (m-3)*(n-3) =2*5 or 5*2
and now u can easily calculate all values of m and n........
