2. If sin a + sin ß + sinr=- 3 Find values of tana/2+ tanß/2 + tanr/2
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Answer:
sin(α+β)=1,sin(α−β)=1/2
⇒α+β=π/2,α−β=π/6
⇒α=
3
π
,β=
6
π
=tan(α+2β)tan(2α+β)
=tan(
3
2π
)tan(
6
5π
)
=(−
3
)(−1/
3
)=1
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