If the tangent drawn at point `(t^2,2t)` on the parabola `y^2=4x` is the same as the normal drawn at point `(sqrt(5)costheta,2sintheta)` on the ellipse `4x^2+5y^2=20,` then (a) `theta=cos^(-1)(-1/(sqrt(5)))` (b) `theta=cos^(-1)(1/(sqrt(5)))` (c) `t=-2/(sqrt(5))` (d) `t=-1/(sqrt(5))`

cos ^(-1)"" (1)/(sqrt(5))+ cos ^(-1) ""(2)/(sqrt(5))=(pi)/(2)

cos^(-1)""(2)/(sqrt(5))+sin ^(-1)""(1)/(sqrt(10))=(pi)/(4)

If the tangents to y^2=4ax at the point (at^2,2at) where |t|>1 is a normal x^2-y^2=a^2 at the point (asectheta,a tan theta) , then t=

cos2theta = (sqrt(2)+1)(costheta -1/sqrt(2))

Simiplify (1)/(7+4sqrt3)+(1)/(2+sqrt5)

Ifint(dx)/(cos^3xsqrt(sin2x))=a(tan^2x+b)sqrt(tanx)+c ,t h e n (a) a=(sqrt(2))/5,b=1/(sqrt(5)) (b) a=(sqrt(2))/5,b=5 (c) a=(sqrt(2))/5,b=-1/(sqrt(5)) (d) a=(sqrt(2))/5, b=sqrt(5)

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)))

Show that 4sin27^0=(5+sqrt(5))^(1/2)+(3-sqrt(5))^(1/2)

Prove that tan^(-1).(1)/(sqrt2) + sin^(-1).(1)/(sqrt5) - cos^(-1).(1)/(sqrt10) = -pi + cot^(-1) ((1 + sqrt2)/(1 - sqrt2))

Find the eccentric angle of a point on the ellipse (x^2)/6+(y^2)/2=1 whose distance from the center of the ellipse is sqrt(5)