Question : If $\tan x = \frac{7}{5}$, the value of $\frac{9 \sin x – \frac{42}{5} \cos x}{15 \sin x + 21 \cos x}$ is:
Option 1: 0
Option 2: 1
Option 3: 0.1
Option 4: 0.5
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Correct Answer: 0.1
Solution :
Given: $\tan \ x = \frac{7}{5}$
Now, $\frac{9 \sin\ x – \frac{42}{5} \cos \ x}{15 \sin\ x + 21 \cos\ x}$
$=\frac{9 \tan \ x – \frac{42}{5} }{15 \tan\ x + 21 }$ [Dividing the expression's numerator and denominator by $\cos x$]
$=\frac{9 \times\frac{7}{5} – \frac{42}{5} }{15 \times\frac{7}{5} + 21 }$
$=\frac{\frac{63 – 42}{5}}{21 + 21}$
$=\frac{\frac{21}{5}}{42}$
$={\frac{1}{10}}$
$ = 0.1$
Hence, the correct answer is 0.1.
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