Asked by Anonymous
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(x-y) = cos2y - cos2x
My Attempt:
RS:
= 1 - 2sin^2y - 1 - 2sin^2x
= 1 - 1 - 2sin^2y - 2sin^2x
= -2sin^2y - 2sin^2x
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(x-y) = cos2y - cos2x
My Attempt:
RS:
= 1 - 2sin^2y - 1 - 2sin^2x
= 1 - 1 - 2sin^2y - 2sin^2x
= -2sin^2y - 2sin^2x
Answers
Answered by
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
LS:
= 2sinxcosx /1 - (1 - 2sin^2x)
= 2sinxcosx / 1 - 1 + 2sin^2x
= 2sinxcosx / 2sin^2x
= cosx / sinx
= cotx
Answered by
Npgm
sin2x / 1 - cos2x = cotx
2sinxcosx / 1 - (1-2sin^2x)
2sinxcosx / 2sin^2x
cosx/sinx = cotx
2sinxcosx / 1 - (1-2sin^2x)
2sinxcosx / 2sin^2x
cosx/sinx = cotx
Answered by
Anonymous
I know how to solve the first question.
Answered by
Npgm
2sin(x+y)sin(x-y) = 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[(1-cos^2x)cos^2y – (1-cos^2y)cos^2x]
2[cos^2y-cos^2xcos^2y – cos^2x + cos^2xcos^2y]
2[cos^2y-cos^2x]
rhs
2 cos^2y - 1 - 2cos^2x+1
2[cos^2y – cos^2x]
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[(1-cos^2x)cos^2y – (1-cos^2y)cos^2x]
2[cos^2y-cos^2xcos^2y – cos^2x + cos^2xcos^2y]
2[cos^2y-cos^2x]
rhs
2 cos^2y - 1 - 2cos^2x+1
2[cos^2y – cos^2x]
Answered by
Anonymous
thanks
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.