`5/12+9/12-2/12-3/12+4/12`

If f_k(x)=1/k(sin^kx+cos^kx) then f_4(x)-f_6 (x)= (A) 1/12 (B) 5/12 (C) (-1)/12 (D) -5/12

Compute: 5/(12)-7/(12)+(11)/(12)

Simplify: 2/3+1/6-2/9 (ii) 12-3 1/2 (iii) 7 5/6-4 3/8+2 7/(12)

Using determinant, find k if area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). a) 12 b) 2 c) -12,-2 d) 12,-2

Replace * by in each of the following to make the statement true: (i) (-6)+(-9) * (-6)-(-9) (ii) (-12)-(-12) * (-12)+(-12)

If z=sqrt(3)-2+i , then principal value of argument z is (where i=sqrt(-1) (1) -(5pi)/(12) (2) pi/(12) (3) (7pi)/(12) (4) (5pi)/(12)

In Fig. , A D=4c m , B D=3c m and C B=12 c m , find cottheta . (a) (12)/5 (b) 5/(12) (c) (13)/(12) (d) (12)/(13) (

In Figure, A D=4c m , B D=3c m and C B=12 c m , find cottheta . (a) (12)/5 (b) 5/(12) (c) (13)/(12) (d) (12)/(13)

(5 1/4-3 1/3) = (a)(12/23) (b) 2 (c) (1 11/12) (d) (11/12)

Factorise : 9a^(2) + (1)/(9a^(2)) - 2-12a + (4)/(3a)