If α and β are two real values of x which satisfy the equation $\sin^{-1} \text{x} + \sin^{-1} (1−\text{x})=\cos^{−1}\text{x}, $then:
$ \alpha + \beta = \frac{1}{2}$
$ \alpha - \beta = \frac{1}{2}$
$ \alpha - \beta = 0$
$ \alpha - \beta = 6$
Answer
$ \alpha + \beta = \frac{1}{2}$
Solution:
$ \sin^{-1}(\text{x})+\sin^{-1}(1-\text{x})=\cos^{-1}(\text{x})$
Now, x ϵ [-1, 1]
and -1 ≤ (1 - x) ≤ 1
x ϵ [0, 2]
Hence, x ϵ [0, 1]
Now we get two solutions of the above equation as 0 and $ \frac{1}{2}$
Thus, the two roots are
x = 0 and $ \text{x}=\frac{1}{2}$
$ α+β=\frac{2}{1}$
and $ \alpha\beta = 0$