Let f(x) =≤ft […] .. If f(x) is continuous in (nπ, (n + 1)π), then

LHL =$\underset{x→(nπ+π/2)^-}{\lim}f(x)=\underset{h→0}{\lim}f(nπ+π/2-h)$

$=\underset{h→0}{\lim}(1+|\cos (nπ + π/2 – h)|)^{ab/|\cos[nπ+π/2–h]|}$

$=\underset{h→0}{\lim}(1 + |\sin (nπ – h)|)^{ab/|\sin(nπ– h)|}$

$=\underset{h→0}{\lim}(1+ | (-1)^n \sin h|)^{ab|1(-1)^n \sin h|}$

$=\underset{h→0}{\lim}(1 + \sin h)^{ab/\sin h}= e^{ab}$

$\left[∵\underset{h→0}{\lim}(1+h)^{h/η}=e^λ\right]$

$V.F. = f(nπ + π/2) = e^a. e^b$

RHL =$\underset{x→(nπ+π/2)^+}{\lim}f(x)=\underset{h→0}{\lim}f(nπ+\frac{π}{2}+h)$

$=\underset{h→0}{\lim}e^{\cot(2nπ + π +2h)/\cot(8nπ + 4π + 8h)}$

$=e^{\underset{h→0}{\lim}(\cot 2h/\cot 8h)}=e^{\underset{h→0}{\lim}\tan 8h / \tan 2h}$

$=e^{4\underset{h→0}{\lim}\tan 8h/8h.\underset{h→0}{\lim}2h/ \tan 2h}=e^{4.1.1} = e^4$

∵ f(x) is continuous in (nπ, (n + 1) π)

∴ LHL = RHL = V.F. $e^{ab} = e^4 = e^{a+b}$

$⇒ ab = 4 = a + b ⇒ a = b = 2$