Asked by Anonymous
Prove that sin^2(Omega) - Cos^2(Omega) / tan(Omega) sin(Omega) + cos(Omega) tan(Omega) = cos(Omega) - cot (Omega) cos (omega)
Answers
Answered by
Reiny
I substituted any angle in the equation the way you typed it, and the equation was false.
I then tried it as
(sin^2 Ø - cos^2 Ø)/(tanØsinØ + cosØtanØ)
using Ø instead of omega for easier typing.
and got
LS = (sinØ+cosØ)(sinØ-cosØ)/(tanØ(sinØ+cosØ)
= (sinØ - cosØ)/(sinØ/cosØ)
= cosØ(sinØ - cosØ)/sinØ
= cosØsinØ/sinØ - cos^2 Ø/sinØ
= cosØ - (cosØ/sinØ)cos‚
= cosØ - cotØcosØ
= RS
I then tried it as
(sin^2 Ø - cos^2 Ø)/(tanØsinØ + cosØtanØ)
using Ø instead of omega for easier typing.
and got
LS = (sinØ+cosØ)(sinØ-cosØ)/(tanØ(sinØ+cosØ)
= (sinØ - cosØ)/(sinØ/cosØ)
= cosØ(sinØ - cosØ)/sinØ
= cosØsinØ/sinØ - cos^2 Ø/sinØ
= cosØ - (cosØ/sinØ)cos‚
= cosØ - cotØcosØ
= RS
Answered by
Anonymous
What does LS and RS stand for?
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