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limx→0{sin2(π2−ax)}sec2(π2−bx)\large \lim_{x\rightarrow 0} \left\{\sin^{2}\left(\dfrac{\pi}{2-ax}\right)\right\}^{\sec^{2}\left(\frac{\pi}{2-bx}\right)}x→0lim{sin2(2−axπ)}sec2(2−bxπ)
If the value of the above limit is in the form ek(a2b2)e^{k\left(\frac{a^2}{b^2}\right)}ek(b2a2), where kkk is a integer, then find −2017k-2017k−2017k.
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