IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

vimalarao

if f(x) = tanx -tanx tanx tanx +tanx tanx tanx tanx tanx+..................+ infinity

0 less than x less than 45 degrees then

_{[0 ]}

^{[/4 ]} f(x) dx =

akhil_o

1)
L1=2x-y=1 bisects line L2=y=mx+c and L3=y=x
let angle between L3 and L2=@

tan @= m2-m1/1+m1m2
tan @=(2-1)/(1+2)=1/3

now angle between L1 and L2 also =@

tan@=m-2/1+2m
1/3=(m-2)/(1+2m)
1+2m=3m-6
m=7

so y=7x+c
but the lines x=y and 2x-y=1 intersect at (1,1)
so y=7x+c passes through (1,1)
1=7+c
c=-6

so equation of line is
y=7x-6

budokai_tenkaichi

roots of equation
(x+1)^1/2-(x-1)^1/2=(4x-1)^1/2

Ronaldooo

Hey,budokai.. whatever
Dont post ur questions in any thread.U did the same thing with my 0*inf.= -1 thread.Dont try to save ur nickels.They are meant to be used.

joyfrancis

is it tanx - tanx tanx tanx + tanx tanx tanx tanx tanx+..................+ infinity.. or

tanx + tanx tanx tanx + tanx tanx tanx tanx tanx+..................+ infinity

anandghegde

@ budokai

edison

Q) f(x) = tanx - tanx tanx tanx +tanx tanx tanx tanx tanx+..................+ infinity

0 less than x less than 45 degrees then

_{[0 ]}

^{[/4 ]} f(x) dx = ?

Sol) Let y =f(x) = tanx - tanx tanx tanx +tanx tanx tanx tanx tanx+...............

Thus y = tanx -tan² x ( tanx - tanx tanx tanx +tanx tanx tanx tanx tanx+....)

or y = tanx - y tan² x

or y + y tan² x = tanx

or y = tanx / (1 + tan² x)

or y = sinx cosx = (sin2x)/2

Thus, I = _{[0 ]}^{[/4 ]} f(x) dx

or I =_{[0 ]}^{[/4 ]} y dx

or I = _{[0 ]}^{[/4 ]} sin2x / 2 dx

or I = -(1/4)(cos

/2 - cos0)

= -(1/4) (0-1) = 1/4

Thus, _{[0 ]}^{[/4 ]} f(x) dx = 1/4

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