I HAVE SOME DOUBT IN THIS SOME
plz solve this sum :- Find the equations of tangents to the curve y= cos (x+y ) , - 2n < x < 2n that are parallel to the line x+2y=0.
Let the point of contact of the tangent be (x 1, y 1) then
(x 1, y 1) lies on y = cos (x + y)
∴ y 1 = cos (x 1 + y 1) ... (1)
Since the tangents are parallel to the line x + 2y = 0
Therefore,
Slope of the tangent of (x 1, y 1) = Slope of the line x + 2y = 0
The equation of the curve is
y = cos (x + y)
Differentiating w.r.t. x, we get
Squaring (1) and (2) and then adding, we get
Putting y 1 = 0 in (1) and (2) we get
Hence the point of contact are 
∴ equation of tangents are
value of
i think value of x1 will be n/2 & 3n/2 not -3n/2 please clarify
The value of x1 will be π/2 and -3π/2.
The value will not be 3π/2 as you can see above that sinx1 = 1. Therefore, for sinx1 to be 1, x1 will have to
be -3π/2 and if x1= 3π/2 , the value of sinx1 will come as -1.