`int_(sintheta)^(costheta)f(xtantheta)dx(w h e r etheta!=(npi)/2,n in I))i se q u a lto` (a) `-costhetaint_1^(tantheta)f(xsintheta)dx` (b)`-tanthetaint_(costheta)^(sintheta)f(x)dx` (c)`sinthetathetaint_1^(tantheta)f(xcostheta)dx` (d)`1/(tantheta)thetaint_(sintheta)^(sinthetatantheta)f(x)dx`

sintheta/(1-cottheta)+costheta/(1-tantheta)= (a) theta (b)1 (c) costheta-sintheta (d) costheta+sintheta

The value of tan^(-1)((xcostheta)/(1-xsintheta))-cot^(-1)((costheta)/(x-sintheta))i s 2theta (b) theta (c) theta/2 (d) independent of theta

Let f(theta)=sin(tan^(-1)((sintheta)/(sqrt(cos2theta)))) ,w h e r e -pi/4ltthetaltpi /4 then the value of d/ (d(tantheta)) f(theta) is

If x=(2sintheta)/(1+costheta+sintheta),t h e n(1-costheta+sintheta)/(1+sintheta) is equal to 1+x (b) 1-x (c) x (d) 1/x

If (acostheta_1,asintheta_1),(acostheta_2,asintheta_2) and (acostheta_3,asintheta_3) represent the vertices of an equilateral triangle inscribed in a circle, then (a) costheta_1+costheta_2+costheta_3=0 (b) sintheta_1+sintheta_2+sintheta_3=0 (c) tantheta_1+tantheta_2+tantheta_3=0 (d) cottheta_1+cottheta_2+cottheta_3=0

If x=a(theta+sintheta), y=a(1+costheta) " then " (dy)/(dx) is

If 5tantheta=4, then (5sintheta-3costheta)/(5sintheta+2costheta) is equal to 0 (b) 1 (c) 1/6 (d) 6

int_(a)^(b) f(x) dx =

Prove that sum_(r=1)^n(1/(costheta+"cos"(2r+1)theta))=(sinntheta)/(2sintheta* costheta*cos (n+1)theta),(w h e r e n in N)dot

If f(x) is a function satisfying f(1/x)+x^2f(x)=0 for all nonzero x , then evaluate int_(sintheta)^(cos e ctheta)f(x)dx