Question #1b1a9

1 Answer
Dec 5, 2017

Explanation below

Explanation:

#2arctan(sqrt(1+x^2)-x)+arctanx=y#

Set #u=arctan(sqrt(1+x^2)-x)#, so #tanu=sqrt(1+x^2)-x#

After using #tan2u=(2tanu)/[1-(tanu)^2]# identity,

#tan2u=(2tanu)/[1-(tanu)^2]#

=#2*(sqrt(1+x^2)-x)/[1-(sqrt(1+x^2)-x)^2]#

=#2*(sqrt(1+x^2)-x)/[2xsqrt(1+x^2)-2x^2]#

=#2*(sqrt(1+x^2)-x)/(2x*[sqrt(1+x^2)-x])#

=#1/x#, hence #2arctan(sqrt(1+x^2)-x)=arctan(1/x)#

Thus,

#y=2arctan(sqrt(1+x^2)-x)+arctanx#

#y=arctan(1/x)+arctanx#

#y=pi/2#