#1+sinx+sin^2x+sin^3+...=1/(1-sinx)#????

Question

#1+sinx+sin^2x+sin^3+...=1/(1-sinx)#
Given #0 < x < pi/2#

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Answer 1 · 2018-06-25T14:02:10

Given #0 < x < pi/2#
So
#0 < sinx < 1#

So LHS is an infinite GP series with first term #a=1# and common ratio #r=sinx <1#. Let sum be #S#

So

#S=1+r+r^2+r^3+.......oo#
#=>S= 1+r(1++r^2+r3+.....oo)#

#=>S= 1+rS#

#=>S(1-r)=1#

#=>S=1/(1-r)=1/(1-sinx)#

So sum of the series is

#1/(1-sinx)#