# find the equation of the line perpendicular to the curve y= (tanx)/(1+tanx) at x= pi/4 ( we can write in terms of pi )

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Given that y= tanx/ (1+ tanx)

We need to find the line perpendicular to the tangent line of y at x= pi/4

Then we will find the tangent point.

==> x= pi/4 ==> y= tanpi/4 / (1+ tanpi/4) = 1/ (1+1) = 1/2

Then the tangent point is ( pi/4, 1/2)

Now we need to find the slope.

We will find the slope of the tangent line first.

Let us differentiate y.

==> y' = ( tanx)'(tanx+1) - (tanx+1)'*tanx / (tanx+1)^2

==> y' = ( sec^2 x (tanx+1) - sec^2 x*tanx / (tanx+1)^2

Now we will subsitute with x= pi/4 to find the slope.

==> y'(pi/4) = [2 (1+1) - 2*1]/ (1+1)^2

==> y' (pi/4) = ( 4 -2)/4 = 2/4 = 1/2

Then the slope of the tangent line if 1/2

Then the slope of the perpendicular line is -2.

Now we will find the equation.

==> y-y1 = m(x-x1)

==> y- 1/2 = -2 ( x-pi/4)

==> y= -2x + pi/2 + 1/2

**==> y= -2x + (pi+1)/2**