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Prove that $\log\sec x=\dfrac {x^2}2+\dfrac {x^4}{12}+\dfrac {x^6}{45}+ --------$
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written 7.3 years ago by |
L.H.S.
$=\log(\sec x ) \\ =\log ( \dfrac 1\cos x ) \\ = \log(\cos x )^{-1} \\ = -1 \log\cos x \\ =-1 \log[1- \dfrac {x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!}…………)] \\ = (\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!}…………)+ \dfrac12 (\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!}…………)^2+ \dfrac13 (\dfrac{x^2}{2!} - \dfrac{x^4}{4!}…………)^3+⋯ \\ = \dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac12 (\dfrac{x^2}{2!})^2+ \dfrac{x^6}{6!} -\dfrac12 (2 \dfrac{x^2}{2!} \dfrac{x^4}{4!} + \dfrac13 (\dfrac{x^2}{2!})^3+⋯ ) \\ =\dfrac{x^2}{2} + x^4 (\dfrac {-1}{24} + \dfrac18) + x^6 [\dfrac1{720} - \dfrac1{48} +\dfrac1{24}]+⋯ \\ =\dfrac{x^2}{2} + \dfrac{x^4}{12} + \dfrac{x^6}{45}+⋯ $
=R.H.S. Hence Proved.
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