Question

If 3 tan0 = 4, evaluate 3 sin0 +2cos0
3 sin0-2cos0​

Answer 1

Answer:

3

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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