The complete solution set of $[\sin^{-1}x]>[\cos^{-1}x]$ is The complete solution set of $[\sin^{-1}x]>[\cos^{-1}x]$ is
$(A)[\sin1,1]\hspace{1 cm}(B)[\frac{1}{\sqrt2},1]\hspace{1 cm}(C)(\cos 1,\sin 1)\hspace{1 cm}(D)[0,1]\hspace{1 cm}$
I think its answer should be (B) as $\sin^{-1}x$ and $\cos^{-1}x$ meet at $\frac{1}{\sqrt2}$.Is my answer and approach correct.If not please tell me the right approach.Can this question be solved without graphs?
 A: This is correct, and can be solved without graphing.
Assume $1>x>0$:
$$\arcsin x>\arccos x$$
Since $\cos x$ is decreasing in that interval, applying $\cos$ changes the inequality direction:
$$\cos(\arcsin x)<x$$
$$\sqrt{1-x^2}<x$$
$$1<2x^2$$
$$x>\frac{1}{\sqrt{2}}$$
A: Answer cannot be $(\bf{B}$)  As stated above $x$ should be greater than $\displaystyle \frac{1}{\sqrt{2}}$ such that
it should have been $\displaystyle \left(\frac{1}{\sqrt{2}},1\right].$
A: $\bf{My\; Solution::}$ Given $$\sin^{-1}(x)>\cos^{-1}(x)\;,$$
Here $$\sin^{-1}(x)\;,\cos^{-1}(x)$$ function is defined When $$-1\leq x\leq 1..........................(1)$$
Now Using The formula
$$\displaystyle \bullet \sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}\Rightarrow \cos^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(x)$$
Put into $$\sin^{-1}(x)>\cos^{-1}(x)\;,$$ We get $\displaystyle \sin^{-1}(x)>\frac{\pi}{2}-\sin^{-1}(x)\;,$
So We get $$\displaystyle 2\sin^{-1}(x)>\frac{\pi}{2}\Rightarrow \sin^{-1}(x)>\frac{\pi}{4}\Rightarrow x>\frac{1}{\sqrt{2}}............(2)$$
From $(1)$ and $(2)\;,$ We Get Common Solution is $$\displaystyle \frac{1}{\sqrt{2}}<x\leq 1\Rightarrow x\in \left(\frac{1}{\sqrt{2}},1\right]$$
