2tan^(-1)x=sin^(-1)(2x)/(1+x^(2))=

Transformations of inverse trigonometric functions need to be handled with care. Consider the identity sin2theta=(2tantheta)/(1+tan^2theta) in its domain of definition. Suppose we set tantheta=x , we have sin2theta=(2x)/(1+x^2) Taking sin^(-1) of both sides yields 2theta=sin^(-1).(2x)/1+x^2i.e.,2tan^(-1)x=sin^(-1).(2x)/(1+ x^2) . But we will discover that the above identity is not valid for all x. Choose x=sqrt3, LHS =2tan^(-1)sqrt3=2xxpi/3=(2pi)/3, RHS=sin.(2sqrt3)/(1+3)=sin^(-1).(sqrt3)/2=pi/3 . And so left hand and right hand side don't match. The reason is that we have disregarded the principal values of inverse functions. So it is well to remember that the iverse trigonometric formmulae have restrictions attached to the argument. When the values of x lie outside the interval of validity then the formula needs to be corrected. Let f(x)=sin^(-1)(x)/(1+x^2),g(x)=2tan^(-1)x . Then the largest interval in R on which f and g both are agree

If xge1 , " then :" 2 tan^(-1)x+sin^(-1)((2x)/(1+x^(2)))=...

Prove that 2tan^(-1)1/x=sin^(-1)((2x)/(x^(2)+1))

If x >1 , then 2\ tan^(-1)x+sin^(-1)((2x)/(1+x^2)) is equal to 4tan^(-1)x (b) 0 (c) pi/2 (d) pi

If x >1 , then 2\ tan^(-1)x+sin^(-1)((2x)/(1+x^2)) is equal to (a) 4tan^(-1)x (b) 0 (c) pi/2 (d) pi