Question

sin3A/cosA + cos3A/sinA =2cotA

Answer 1

L.H.S. = cos3A/sinA + sin3A /cosA

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

= 1/sinA cosA[ 4 cos^4A - 3 cos^2A + 3 sin^2A - 4 sin^4A]

=2/2sinA cosA[4( cos^2A- sin^2A)(cos^2A+sin^2A)-3( cos^2A-sin^2A)]

= 2/sin2A ( cos^2A - sin^2A)[4(cos^2A+ sin^2A) - 3 ]

= 2 cos2A/sin2A *[4-3]

= 2 cot2A = R. H. S