3cot(θ)=2 Solve for a where a =2sin(θ)+3cos(θ) / 2sin(θ) −3cos(θ)
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Answer:
Step-by-step explanation:
Given,
3cotθ=2
cotθ=
3
2
tanθ=
2
3
2sinθ+6cosθ
4sinθ−3cosθ
=
2sinθ+6cosθ
4sinθ−3cosθ
×
sinθ
sinθ
---------- ( Multiply and divide by sinθ )
=
sinθ
2sinθ+6cosθ
sinθ
4sinθ−3cosθ
=
2+6(
sinθ
cosθ
)
4−3(
sinθ
cosθ
)
=
2+6cotθ
4−3cotθ
=
2+6×
3
2
4−3×
3
2
=
2+4
4−2
=
6
2
=
3
1
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