Question 71
If A lies in the first quadrant and $$6 \tan A = 5$$, then the value of $$\frac{8 \sin A - 4 \cos A}{\cos A + 2 \sin A}$$ is:
Solution
$$6 \tan A = 5$$
$$\tan A = \frac{5}{6}$$
$$\frac{8 \sin A - 4 \cos A}{\cos A + 2 \sin A}$$
= $$\frac{cos A(8 \tan A - 4)}{\cos A(1 + 2 \tan A)}$$
= $$\frac{8 \tan A - 4}{1 + 2 \tan A}$$
Put the value of $$\tan \theta$$,
= $$\frac{8 \times \frac{5}{6} - 4}{1 + 2 \times \frac{5}{6}}$$
= $$\frac{\frac{20}{3} - 4}{1 + \frac{5}{3}}$$
= $$\frac{\frac{8}{3}}{\frac{8}{3}}$$ = 1
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