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

Options

A. `[-1,1]`
B. `(-oo,-1]`
C. `[1,oo)`
D. `(-oo,1)`

Answer

To solve the problem, we need to determine the largest interval in \( \mathbb{R} \) on which the functions \( f(x) = \frac{\sin^{-1}(x)}{1 + x^2} \) and \( g(x) = 2\tan^{-1}(x) \) agree. This involves finding the domains of both functions and then determining their intersection.

### Step-by-step Solution:

1. **Identify the function \( f(x) \)**:
\[ f(x) = \frac{\sin^{-1}(x)}{1 + x^2} \]

2. **Determine the domain of \( f(x) \)**:
- The function \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\).
- The denominator \( 1 + x^2 \) is always positive for all \( x \) in \( \mathbb{R} \).
- Therefore, the domain of \( f(x) \) is:
\[ D_f = [-1, 1] \]

3. **Identify the function \( g(x) \)**:
\[ g(x) = 2\tan^{-1}(x) \]

4. **Determine the domain of \( g(x) \)**:
- The function \( \tan^{-1}(x) \) is defined for all \( x \) in \( \mathbb{R} \).
- Therefore, the domain of \( g(x) \) is:
\[ D_g = \mathbb{R} \]

5. **Find the intersection of the domains**:
- We need to find the intersection of \( D_f \) and \( D_g \):
\[ D_f \cap D_g = [-1, 1] \cap \mathbb{R} = [-1, 1] \]

6. **Conclusion**:
- The largest interval in \( \mathbb{R} \) on which both functions agree is:
\[ [-1, 1] \]

### Final Answer:
The largest interval in \( \mathbb{R} \) on which \( f(x) \) and \( g(x) \) both agree is \( [-1, 1] \).