If 4 tan theta = 3, then left(frac{4 sin thet | vedclass

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(D) Given,$4 	an 	heta = 3$$\Rightarrow 	an 	heta = \frac{3}{4}$ ..........$(i)$To evaluate the expression $\frac{4 \sin 	heta - \cos 	heta}{4 \

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$\frac{1}{2}$

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(D) Given,$4 	an 	heta = 3$$\Rightarrow 	an 	heta = \frac{3}{4}$ ..........$(i)$To evaluate the expression $\frac{4 \sin 	heta - \cos 	heta}{4 \sin 	heta + \cos 	heta}$,divide both the numerator and the denominator by $\cos 	heta$:$= \frac{\frac{4 \sin 	heta}{\cos 	heta} - \frac{\cos 	heta}{\cos 	heta}}{\frac{4 \sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\cos 	heta}}$$= \frac{4 	an 	heta - 1}{4 	an 	heta + 1}$Now,substitute the value of $	an 	heta = \frac{3}{4}$ from equation $(i)$:$= \frac{4(\frac{3}{4}) - 1}{4(\frac{3}{4}) + 1}$$= \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$.

Part $I$	Part $II$
$1.$ $\cos(90^\circ - \theta)$	$a.$ $\sec \theta$
$2.$ $\cot(90^\circ - \theta)$	$b.$ $\sin \theta$
$3.$ $\operatorname{cosec}(90^\circ - \theta)$	$c.$ $1$
	$d.$ $\tan \theta$

If $4 \tan \theta = 3,$ then $\left(\frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta}\right)$ is equal to

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Which of the following groups truly matches the data of Part $I$ with the data of Part $II$?

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