Question 74

If $$5cot\theta=3$$, find the value of $$\frac{6sin\theta-3cos\theta}{7sin\theta+3cos\theta}$$.

Solution

$$5cot\theta=3$$

$$cot\theta=3/5$$

$$\frac{6sin\theta-3cos\theta}{7sin\theta+3cos\theta}$$

= $$\frac{sin\theta(6-\frac{3cos\theta}{sin\theta})}{sin\theta(7 + \frac{3cos\theta}{sin\theta})}$$

= $$\frac{6 - 3cot\theta}{7 + 3cot\theta}$$

Put the value of cot$$\theta$$,

= $$\frac{6 - 3 \times \frac{3}{5}}{7 + 3\times \frac{3}{5}}$$

= $$\frac{6 -\frac{9}{5}}{7 + \frac{9}{5}}$$

=$$\frac{21}{44}$$

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