Question #a336e

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Answer 1 · 2016-12-18T15:36:01

#LHS=(cos3A+sin3A)/(cosA-sinA)#

#=(4cos^3A-3cosA+3sinA-4sin^3A)/(cosA-sinA)#

#=(4(cos^3A-sin^3A)-3(cosA-sinA))/(cosA-sinA)#

#=(4(cos^3A-sin^3A))/(cosA-sinA)-3#

#=(4(cosA-sinA)(cos^2A+cosAsinA+sin^2A))/(cosA-sinA)-3#

#=4(1+cosAsinA)-3#

#=4+4cosAsinA-3#

#=1+2*2cosAsinA#

#=1+2sin2A=RHS#

Proved