Which of the following is the value of `sin 27^0-cos 27^0?` (a) `-(sqrt(3-sqrt(5)))/2` (b) `(sqrt(5-sqrt(5)))/2` (c) `-(sqrt(5)-1)/(2sqrt(2))` (d) none of these

(a) sqrt(175) (b) 2sqrt(112) (c) sqrt(125) (d) 5sqrt(119)

Which of the following is equal to root(3)(-1) a. (sqrt(3)+sqrt(-1))/2 b. (-sqrt(3)+sqrt(-1))/(sqrt(-4)) c. (sqrt(3)-sqrt(-1))/(sqrt(-4)) d. -sqrt(-1)

(2sqrt(-5)+3sqrt(-2))(-3sqrt(-8)-sqrt(-20))

The centroid of an equilateral triangle is (0, 0). If two vertices of the triangle lie on x+y=2sqrt(2), then one of them will have its coordinates. (a) (sqrt(2)+sqrt(6),sqrt(2)-sqrt(6)) (b) (sqrt(2)+sqrt(3),sqrt(2)-sqrt(3)) (c) (sqrt(2)+sqrt(5),sqrt(2)-sqrt(5)) (d) none of these

If a<0,b>0, then sqrt(a)sqrt(b) is equal to (a) -sqrt(|a|b) (b) sqrt(|a|b)i (c) sqrt(|a|b) (d) none of these

Solve sqrt(5)x^(2) + x + sqrt(5)=0

Simplify : sqrt2 - sqrt(18) + sqrt(27) - sqrt(32)

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

Solve (sqrt(5)-1)/(sinx)+(sqrt(10+2sqrt(5)))/(cosx)=8,x in (0,pi/2)