<s>[INST] How to prove that $\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$ by using Mean Value Theorem? Given Question is Prove that 
$$\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$$
when $0<a<b<1$ using Mean Value Thoerem.
I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and used LMVT and established that $\frac{b}{\sqrt{a-c^2}}=\sin^{-1}b$ where $c \in (0,b)$.
But I don't know how to proceed after this. Please guide.
 [/INST] Using $f (x)=arcsin (x) $ in the interval $[a,b] $  we have $\frac {\arcsin (b)-\arcsin (a)}{b-a}=\frac {1}{\sqrt {1-c^2} }$ now we have $a <c <b $ hence $-a^2>-c^2>-b^2$ adding 1 and recipeocating we have $\frac {1}{1-a^2}<\frac {1}{1-c^2}<\frac {1}{1-b^2} $ hence taking root completes the proof.
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