It 3 tan O= 4, then evaluate
Answers
Answer:
3
Step-by-step explanation:
Given
3tan\theta = 43tanθ=4
\implies tan\theta = \frac{4}{3}⟹tanθ=
3
4
----(1)
Now ,
The value of
\frac{3sin\theta+2cos\theta}{2cos\theta-2sin\theta}
2cosθ−2sinθ
3sinθ+2cosθ
Divide numerator and denominator by cos\thetacosθ , we get
=\frac{\frac{3sin\theta}{cos\theta}+\frac{2cos\theta}{cos\theta}}{\frac{3sin\theta}{cos\theta}-\frac{2cos\theta}{cos\theta}}
cosθ
3sinθ
−
cosθ
2cosθ
cosθ
3sinθ
+
cosθ
2cosθ
=\frac{3tan\theta+2}{3tan\theta-2}
3tanθ−2
3tanθ+2
=\frac{3\times\frac{4}{3}+2}{3\times\frac{4}{3}-2}
3×
3
4
−2
3×
3
4
+2
/* from (1)*/
After cancellation, we get
= \frac{4+2}{4-2}
4−2
4+2
= \frac{6}{2}
2
6
= 33
Therefore,
\frac{3sin\theta+2cos\theta}{2cos\theta-2sin\theta}
2cosθ−2sinθ
3sinθ+2cosθ
= 3
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