Let #arcsinx=A, &, arctan2x=B#, so that, #sinA=x, &, tanB=2x, A in [-pi/2,pi/2], B in (-pi/2,pi/2)#
We will consider the case #A, B in [0,pi/2]#, so that all trigonometric ratios will be #+ve#. The other case can be dealt with similarly.
Now reqd. value #=cos(A-B)=cosAcosB+sinAsinB.......(I)#
#sinA=x rArr cosA =sqrt(1-x^2)#
#tanB=2x rArr sec^2B=1+tan^2B=1+4x^2rArr secB=sqrt(1+4x^2) rArr cosB=1/sqrt(1+4x^2)#
#tanB=2xrArrcotB=1/(2x)rArrcsc^2B=1+cot^2B=1+1/(4x^2)=(4x^2+1)/(4x^2)rArrcscB=sqrt(1+4x^2)/(2x)rArrsinB=(2x)/sqrt(1+4x^2)#
Sub.ing all these in #(I)#, we have,
The Reqd. Value#=sqrt(1-x^2)/sqrt(1+4x^2)+(2x^2)/sqrt(1+4x^2)#
#={sqrt(1-x^2)+2x^2}/sqrt(1+4x^2)#
Hope, this is Helpful! Enjoy Maths. , &, yes, don't forget to consider
the case #A,B in [-pi/2,0] !#
