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8. Introduction to Trigonometry
medium
$\tan \theta=\frac{4}{3},$ તો $\frac{5 \sin \theta+2 \cos \theta}{3 \sin \theta-\cos \theta}=\ldots \ldots \ldots \ldots$
A
$\frac{22}{13}$
B
$2$
C
$\frac{26}{9}$
D
$\frac{7}{2}$
Solution
$\frac{5 \sin \theta+2 \cos \theta}{3 \sin \theta-\cos \theta}=\frac{\frac{5 \sin \theta}{\cos \theta}+\frac{2 \cos \theta}{\cos \theta}}{\frac{3 \sin \theta}{\cos \theta}-\frac{\cos \theta}{\cos \theta}}$
($\because $ Dividing each term of numerator and denominator by $\cos \theta, \cos \theta \neq 0$ )
$=\frac{5 \tan \theta+2}{3 \tan \theta-1}=\frac{5\left(\frac{4}{3}\right)+2}{3\left(\frac{4}{3}\right)-1}=\frac{\frac{20}{3}+2}{4-1}=\frac{20+6}{3 \times 3}=\frac{26}{9}$
Standard 10
Mathematics