Tangents drawn from `(2, 0)` to the circle `x^2 + y^2 = 1` touch the circle at `A and B`. Then (A) `A-= (1/2, sqrt(3)/2)`, B -= (-1/2, - sqrt(3)/2)` (B) `A-= (- 1/2, (-sqrt(3)/2)` , B -= (1/2, sqrt(3)/2)` (C) `A-= (1/2, sqrt(3)/2)`, B-= (1/2, - sqrt(3)/2)`` (D) `A-= (1/2, - sqrt(3)/2), B-= (1/2, sqrt(3)/2)`

Sin (A + B) = 1/(sqrt2) Cos(A - B) = (sqrt3)/2

((2+sqrt(3))/2)/(1+(sqrt(3)+1)/2) + ((2-sqrt(3))/2)/(1-((sqrt(3)-1)/2))

Evaluate : Find the value of sqrt( 3 + sqrt(5)) a ) sqrt(3)/2 + 1/sqrt(2) b ) sqrt(3)/2 - 1/2 c ) sqrt(5)/2 - 1/sqrt(2) d ) sqrt(5)/sqrt(2) + 1/sqrt(2)

If a = (1)/(2 - sqrt(3)) , b = (1)/(2 + sqrt(3)) , find the value of ((a + b)/(a -b))^(2) .

The value of sin(1/4sin^(-1)(sqrt(63))/8) is 1/(sqrt(2)) (b) 1/(sqrt(3)) (c) 1/(2sqrt(2)) (d) 1/(3sqrt(3))

(1)/(sqrt(9)-sqrt(8)) is equal to: 3+2sqrt(2)(b)(1)/(3+2sqrt(2)) (c) 3-2sqrt(2)(d)(3)/(2)-sqrt(2)

Let A=[[cos theta,-sin theta],[sin theta,cos theta]] Find the value of A^(-50) at theta=(pi)/2 (A) [[-(sqrt(3))/(2),-(1)/(2)],[-(1)/(2),(sqrt(3))/(2)]] (B) [[(sqrt(3))/(2),(1)/(2)],[(-1)/(2),(sqrt(3))/(2)]] (C) [[-(sqrt(3))/(2),(1)/(2)],[(1)/(2),(sqrt(3))/(2)]] (D) [[(1)/(2),(sqrt(3))/(2)],[(sqrt(3))/(2),(-1)/(2)]]

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)