1357again see your graph in 3rd quadrant
I have doubt on that
24thats what I did ......
why mine is wrong then ??
1but sir graph is coming in all four quadrants and it is becoming very complex...........especially for 3rd and 4th quadrants
1357
Manish Shankar
·2009-10-03 08:39:04
yes it is correct for first quadrant
1357
Manish Shankar
·2009-10-03 06:26:12
just complete the graph so that I can close the thread
24
eureka123
·2009-10-03 00:11:14
sir plz telll me if I have made mistake
24
eureka123
·2009-10-02 23:35:35
Isnt this one also same as here ??
http://targetiit.com/iit-jee-forum/posts/daily-graph-1-08-09-11530.html
it simplfies to tan({x}{y})=1
=>{x}{y}=nπ+π4=m where m is variable of course...
now for diff values of m there will be rectangular hyperbolas..
1357
Manish Shankar
·2009-10-02 23:52:15
eureka
{x}{y}=nπ+π/4
ans {x}{y}<1 so what is the value of n
24
eureka123
·2009-10-02 23:52:11
yep karna is rite..graph only this part only
24
eureka123
·2009-10-02 23:44:37
it will be in 1 and 3rd quad only becoz tan gioves positive value in htese only
1357
Manish Shankar
·2009-10-02 23:41:18
yeah its kind of repeated
m in your solution sound of no use.
can you simplify it further
24
eureka123
·2009-10-02 23:37:29
Some discontinuity too.....
I ma working on quadrants...
