If tan (theta 1 + theta 2) = √3 and sec (theta 1 -theta 2) = 2/root 3
then finde sin 2theta 1 + tan 3 theta 2
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Answer:
tan
2
θ+2
3
tanθ=1
tan
2
θ+2
3
tanθ−1=0
Add +4 on both sides
tan
2
θ+2
3
tanθ+3=4
(tanθ+
3
)
2
=4
tanθ+
3
=±2
tanθ=2−
3
, tanθ=−(2+
3
)
tan(
12
π
) tan(
2
−π
)
x=nπ±
12
π
for n=0,1,2…
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