Let `(3pi)/4 < theta < pi` and `sqrt(2 cot theta+1/sin^2 theta) = k - cot theta` then `k=`

f(x)=min(sinx, cosx),x in (-pi,pi) the values of x where f(x) is non differentiable, is the set A . Then A is subset of (A) {(-3pi)/4, (-pi)/4, (3pi)/4} (B) {(-pi)/4, pi/4, (3pi)/3} (C) {(-3pi)/4, (3pi)/4} (D) {(-3pi)/4, (-pi)/4, pi/4, (3pi)/4}

Let the domain and range of inverse circular functions are defined as follows: sin^(-1)xlt(3pi)/4 then solution of x is

The possible values of theta in (0,pi) such that sin (theta) + sin (4theta) + sin(7theta) = 0 are (1) (2pi)/9 , i/4 , (4pi)/9, pi/2, (3pi)/4 , (8pi)/9 (2) pi/4, (5pi)/12, pi/2 , (2pi)/3, (3pi)/4, (8pi)/9 (3) (2pi)/9, pi/4 , pi/2 , (2pi)/3 , (3pi)/4 , (35pi)/36 (4) (2pi)/9, pi/4, pi/2 , (2pi)/3 , (3pi)/4 , (8pi)/9

The possible values of theta in (0,pi) such that sin (theta) + sin (4theta) + sin(7theta) = 0 are (1) (2pi)/9 , i/4 , (4pi)/9, pi/2, (3pi)/4 , (8pi)/9 (2) pi/4, (5pi)/12, pi/2 , (2pi)/3, (3pi)/4, (8pi)/9 (3) (2pi)/9, pi/4 , pi/2 , (2pi)/3 , (3pi)/4 , (35pi)/36 (4) (2pi)/9, pi/4, pi/2 , (2pi)/3 , (3pi)/4 , (8pi)/9

If 2|sin2alpha|=|tanbeta+cotbeta|,alpha,beta in (pi/2,pi), then the value of alpha+beta is (a) (3pi)/4 (b) pi (c) (3pi)/2 (d) (5pi)/4

For f(x)=sin^2x ,x in (0,pi) point of inflection is: pi/6 (b) (2pi)/4 (c) (3pi)/4 (d) (4pi)/3

Let vec a* vec b=0,w h e r e vec aa n d vec b are unit vectors and the unit vector vec c is inclined at an angle theta to both vec aa n d vec bdot If vec c=m vec a+n vec b+p( vec axx vec b),(m ,n , p in R), then a. -pi/4lt=thetalt=pi/4 b. pi/4lt=thetalt=(3pi)/4 c. 0lt=thetalt=pi/4 d. 0lt=thetalt=(3pi)/4

2sin^2 (3pi)/4 +2cos^2 pi/4 +2sec^2 pi/3=10

If cos^2 2x+2cos^2x=1,\ x in (-pi,pi), then x can take the values : +-pi/2 (b) +-pi/4 +-(3pi)/4 (d) none of these

Let theta = tan^(-1) ( tan . (5pi)/4) " and " phi = tan^(-1) ( - tan . (2pi)/3) then