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