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.
 A: According to MVT, there exists $c$, $a\leq c\leq b$ such that 
$$f(b)-f(a)=f'(c)(b-a)$$
so with $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$ we see
$$\sin^{-1}b-\sin^{-1}a=\dfrac{1}{\sqrt{1-c^2}}(b-a)$$
step by step
\begin{align}
a\leq &c\leq b\\
a^2\leq &c^2\leq b^2\\
1-b^2\leq &1-c^2\leq 1-a^2\\
\dfrac{1}{\sqrt{1-a^2}}\leq &\dfrac{1}{\sqrt{1-c^2}}\leq \dfrac{1}{\sqrt{1-b^2}}\\
\dfrac{b-a}{\sqrt{1-a^2}}\leq &\dfrac{b-a}{\sqrt{1-c^2}}\leq \dfrac{b-a}{\sqrt{1-b^2}}\\
\dfrac{b-a}{\sqrt{1-a^2}}\leq &\sin^{-1}b-\sin^{-1}a\leq \dfrac{b-a}{\sqrt{1-b^2}}
\end{align}
A: 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.
A: HINT.- $f$ is increasing and $f(b)-f(a)=f'(\theta)(b-a)$ for $\theta\in ]a,b[$ so you have to prove $\dfrac{1}{\sqrt{1-a^2}}\lt f'(\theta)\lt\dfrac{1}{\sqrt{1-b^2}}$ knowing that $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$.
Prove now that $f'$ is increasing in $0\lt x\lt1$
