" 16."1-(b-c)^(2)

If in a A B C , =a^2-(b-c)^2, then tanA is equal to 15//16 b. 8//15 c.8//17 d. 1//2

If a =2b and b=4c then ((a^(1/4))/(16bc))^(1/2) =

A B C is an isosceles triangle. If the coordinates of the base are B(1,3) and C(-2,7) , the coordinates of vertex A can be (a) (1,6) (b) (-1/2,5) (c) (5/6,6) (d) none of these

If |(1,1,1),(2,b,c),(4,b^(2),c^(2))| and |A| = in [2, 16]. 2, b, c and in A.P. the range of c is

Evaluate: ("Lim")_(x->0)((log)_(secx/2)(cosx))/((log)_(secx)(cos(x//2))) 1 (b) 16 (c) 4 (d) 2

The value of ((2.3)^(3)-.027)/((2.3)^(2)+.69+.09) is (a) 0 (b) 1.6(c)2(d)3.4 is (a)0(b)

Let the numbers 2, b, c be in an AP and A= [{:(1, 1,1), (2, b, c),(4, b^(2), c^(2)):}] If "det"(A) in [2, 16] , then c lies in the interval.

Factorise the using the identity a ^(2) -b ^(2) = (a +b) (a-b). (1)/(36) a ^(2) b ^(2) - (16)/(49) b ^(2) c ^(2)