`(sqrt(1+b)-1)/b`

If (sqrt(3)-\ 1)/(sqrt(3)+\ 1)=a-b\ sqrt(3) , then (a) a=2,\ b=1 (b) a=2,\ b=-1 (c) a=-2,\ b=1 (d) a=b=1

Remove the irrationality in the denominator a. sqrt((sqrt(2)-1)/(sqrt(2)+1)) b. 1/(1+sqrt(2)+sqrt(3))

In each of the following determine rational number a\ a n d\ b : (i) (sqrt(3)-1)/(sqrt(3)+1)=a-bsqrt(3) (ii) (4+\ sqrt(2))/(2+sqrt(2))=a-sqrt(b)

If both a and b are rational numbers, find the values of a and b in each of the following equalities : (sqrt(3)-1)/(sqrt(3)+1)=a+bsqrt(3)

If two different tangents of y^2=4x are the normals to x^2=4b y , then (a) |b|>1/(2sqrt(2)) (b) |b| 1/(sqrt(2)) (d) |b|<1/(sqrt(2))

Suppose a ,b, and c are in A.P. and a^2, b^2 and c^2 are in G.P. If a

Let A={:[(sqrt(3),-1),(2+sqrt(3),1-sqrt(3))]:},B={:[(-sqrt(3),2),(2-sqrt(3),1+sqrt(3))]:} Find A+B.

f(x)=(1/sqrt(b-a))(((sqrt((b-a)/a))sin2x)/sqrt(1+((sqrt(b-a)/a))sinx)^2)(sqrt(a+btan^2x) at x=3pi/4

A(1/(sqrt(2)),1/(sqrt(2))) is a point on the circle x^2+y^2=1 and B is another point on the circle such that are length A B=pi/2 units. Then, the coordinates of B can be (a) (1/(sqrt(2)),-1/sqrt(2)) (b) (-1/(sqrt(2)),1/sqrt(2)) (c) (-1/(sqrt(2)),-1/(sqrt(2))) (d) none of these

Tangents are drawn to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1,(a > b), and the circle x^2+y^2=a^2 at the points where a common ordinate cuts them (on the same side of the x-axis). Then the greatest acute angle between these tangents is given by (A) tan^(-1)((a-b)/(2sqrt(a b))) (B) tan^(-1)((a+b)/(2sqrt(a b))) (C) tan^(-1)((2a b)/(sqrt(a-b))) (D) tan^(-1)((2a b)/(sqrt(a+b)))