If tan theta = frac{4}{3},then frac{5 sin the | vedclass

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(C) Given,$	an 	heta = \frac{4}{3}$.To evaluate the expression $\frac{5 \sin 	heta + 2 \cos 	heta}{3 \sin 	heta - \cos 	heta}$,divide both the n

Vedclass · Accepted Answer

$\frac{26}{9}$

Vedclass · Answer

(C) Given,$	an 	heta = \frac{4}{3}$.To evaluate the expression $\frac{5 \sin 	heta + 2 \cos 	heta}{3 \sin 	heta - \cos 	heta}$,divide both the numerator and the denominator by $\cos 	heta$ (where $\cos 	heta 
eq 0$):$= \frac{\frac{5 \sin 	heta}{\cos 	heta} + \frac{2 \cos 	heta}{\cos 	heta}}{\frac{3 \sin 	heta}{\cos 	heta} - \frac{\cos 	heta}{\cos 	heta}}$$= \frac{5 	an 	heta + 2}{3 	an 	heta - 1}$Substitute $	an 	heta = \frac{4}{3}$ into the expression:$= \frac{5(\frac{4}{3}) + 2}{3(\frac{4}{3}) - 1}$$= \frac{\frac{20}{3} + 2}{4 - 1}$$= \frac{\frac{20 + 6}{3}}{3}$$= \frac{26}{3 	imes 3} = \frac{26}{9}$.

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

If $\tan \theta = \frac{4}{3}$,then $\frac{5 \sin \theta + 2 \cos \theta}{3 \sin \theta - \cos \theta} = \ldots \ldots$

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