Posted by Anonymous on .
Prove:
sin2x / 1  cos2x = cotx
My Attempt:
LS:
= 2sinxcosx /  1  (1  2sin^2x)
= 2sinxcosx /  1 + 2sin^2x
= cosx / sinx  1
= cosx / sinx  1/1
= cosx / sinx  sinx / sinx

Prove:
2sin(x+y)sin(xy) = cos2y  cos2x
My Attempt:
RS:
= 1  2sin^2y  1  2sin^2x
= 1  1  2sin^2y  2sin^2x
= 2sin^2y  2sin^2x

Followup 
Anonymous,
Solved the first problem, I know what I did wrong...
LS:
= 2sinxcosx /1  (1  2sin^2x)
= 2sinxcosx / 1  1 + 2sin^2x
= 2sinxcosx / 2sin^2x
= cosx / sinx
= cotx 
Math  Trig  Double Angles 
Npgm,
sin2x / 1  cos2x = cotx
2sinxcosx / 1  (12sin^2x)
2sinxcosx / 2sin^2x
cosx/sinx = cotx 
Math  Trig  Double Angles 
Anonymous,
I know how to solve the first question.

Math  Trig  Double Angles 
Npgm,
2sin(x+y)sin(xy) = cos2y  cos2x
lhs
2(sinx cosy + cosx siny) (sin x cos y – cosx siny)
2( sin^2xcos^2y – sinxcosycosxsiny + sinxcosycosxsiny – cos^2xsin^2y)
2(sin^2xcos^2y – cos^2xsin^2y)
2[(1cos^2x)cos^2y – (1cos^2y)cos^2x]
2[cos^2ycos^2xcos^2y – cos^2x + cos^2xcos^2y]
2[cos^2ycos^2x]
rhs
2 cos^2y  1  2cos^2x+1
2[cos^2y – cos^2x] 
Math  Trig  Double Angles 
Anonymous,
thanks