`("lim")_(xvec0)((x+y)sec(x+y)-xsecx)/yi se q u a lto` `secx(xtanx+1)` (b) `xtanx+secx` `xsecx+tanx` (d) none of these

("lim")_(xvec0)("cos"(tanx)-cosx)/(x^4)i se q u a lto 1/6 (b) -1/3 (c) 1/2 (d) 1

lim_ (x rarr0) ((x + y) sec (x + y) -y sec y) / (x)

int[lim_(hrarr0)(sec(x+h)-secx)/(h)]dx=

y sec x = tanx+c is a solution of the D.E.

lim _(xto0) ((cos x -secx )/(x ^(2) (x+1)))=

(lim)_(nvecoo)1/x[int_y^a e^(sin^2t))dt-int_(x+y)^a e^(sin^2t))dt-]i se q u a lto (0,1] (b) [1,oo) (0,oo) (d) none of these

int((ln(tan x))/(sin x cos x)dxi se q u a lto (a) (1)/(2)ln(tan x)+c(b)(1)/(2)ln(tan^(2)x)+c(c)(1)/(2)(ln(tan x))^(2)+c(d) none of these

(lim)_(xvec0)((sqrt(1+x s in x)-sqrt(cos2x))/(tan^2(x//2))) is equal to 1/6 b. 6 c.3 d. 2

(lim_(x rarr0)(|s in x|)/(x) is a.1b*-1c.0d .none of these

Differentiate wrt x : (sec x-tanx)/(secx+tanx)