y=(2)/(sqrt(a^(2)-b^(2)))(tan^(-1)(sqrt((a-b)/(a+b))tan(x)/(2)))

A: int (1)/(3+2 cos x)dx=(2)/(sqrt(5))"Tan"^(-1)((1)/(sqrt(5))"tan" (x)/(2))+c R: If a gt b then int (dx)/(a+b cosx)=(2)/(sqrt(a^(2)-b^(2)))Tan^(-1)[(sqrt(a-b))/(a+b)"tan"(x)/(2)]+c

2Tan^(-1)((sqrt(a-b))/(a+b)"tan"x/2)=

(d )/(dx ) { (2)/( sqrt(a ^(2) - b ^(2))) Tan ^(-1) (( sqrt (a -b ))/( a + b) tan (x )/(2 )) }=

(d)/(dx ) { Tan ^(-1) ( sqrt(|(a -b)/(a +b)|)tan ""(x)/(2))}=

int(1)/(a^(2)-b^(2)cos^(2)x)dx(a>b)=(1)/(a sqrt(a^(2)-b^(2)))tan^(-1)[(a tan x)/(sqrt(a^(2)-b^(2)))]+c

y = tan ^ (- 1) [sqrt ((ab) / (a + b)) (tan x) / (2)]

A: int (1)/(3+4cosx)dx=(1)/(sqrt(7))log|(sqrt(7)+tan(x//2))/(sqrt(7)-tan(x//2))|+c R : If a lt b then int(dx)/(a+b cosx)= (1)/(sqrt(b^(2)-a^(2)))log|(sqrt(b+a)+sqrt(b-a)tan(x//2))/(sqrt(b+a)-sqrt(b-a)tan(x//2))|+c

If a,b,c>0 and s=(a+b+c)/(2), prove thattan^(-1)sqrt((2as)/(bc))+tan^(-1)sqrt((2bs)/(ca))+tan^(-1)sqrt((2cs)/(ab))=pi

A: int (1)/(4+5 sinx)dx=(1)/(3)log|(2tan(x//2)+1)/(2tan(x//2)+4)|+c R : If 0 lt a lt b , then int(dx)/(a+b sin x)=(1)/(sqrt(b^(2)-a^(2)))log|(a tan (x//2)+b-sqrt(b^(2)-a^(2)))/(a tan (x//2)+b+sqrt(b^(2)-a^(2)))|+c

Tangents are drawn to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1,(a > b), and the circle x^2+y^2=a^2 at the points where a common ordinate cuts them (on the same side of the x-axis). Then the greatest acute angle between these tangents is given by (A) tan^(-1)((a-b)/(2sqrt(a b))) (B) tan^(-1)((a+b)/(2sqrt(a b))) (C) tan^(-1)((2a b)/(sqrt(a-b))) (D) tan^(-1)((2a b)/(sqrt(a+b)))