Practicing Success
The solution set of the equation $cos^{-1} x - sin^{-1}x = sin^{-1} (1-x), $ is |
[-1, 1] [0, 1/2] [-1, 0] none of these |
none of these |
We observe that the LHS of the given equation is defined for -1 ≤ x ≤ 1, whereas RHS is defined for -1 ≤ (1- x) ≤ 1 i.e 0 ≤ x ≤ 2. $cos^{-1} x - sin^{-1}x = sin^{-1} (1-x)$ $⇒ \frac{\pi}{2} -2 sin^{-1} x = sin^{-1}(1- x)$ $⇒ \frac{\pi}{2}- sin^{-1} (1 - x) = 2 sin^{-1} x $ $⇒ cos^{-1} (1-x) = cos^{-1}(1-2x^2)$ $⇒ 1- x = 1- 2x^2 ⇒2x^2 -x = 0 ⇒ x = 0, \frac{1}{2}$ |