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Question #83c24

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Feb 14, 2017

Given #6tanx-5=0=>tanx=5/6#

Now

#(2sinx-cosx)/(5cosx+9sinx)#

Dividing both numerator and denominator by #cosx# the expression becimemes.

#(2sinx-cosx)/(5cosx+9sinx)#

#=((2sinx)/cosx-cosx/cosx)/((5cosx)/cosx+(9sinx)/cosx)#

#=(2tanx-1)/(5+9tanx)#

#=(2xx5/6-1)/(5+9xx5/6)#

#=(5/3-1)/(5+15/2)#

#=(2/3)/(25/2)=2/3xx2/25=4/75#

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