If theta is in the first quadrant and tan the | vedclass

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(C) Given $	an 	heta = \frac{3}{4}$ and $	heta$ is in the first quadrant.Since $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{3}{4}

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$\frac{11}{8}$

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(C) Given $	an 	heta = \frac{3}{4}$ and $	heta$ is in the first quadrant.Since $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{3}{4}$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.Thus,$\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.Now,simplify the expression using trigonometric identities:$	an \left(\frac{\pi}{2}-	hetaight) = \cot 	heta$$\sin (\pi-	heta) = \sin 	heta$$\sin \left(\frac{3 \pi}{2}+	hetaight) = -\cos 	heta$$\cot (2 \pi-	heta) = -\cot 	heta$Substituting these into the expression:$\frac{\cot 	heta - \sin 	heta}{-\cos 	heta - (-\cot 	heta)} = \frac{\cot 	heta - \sin 	heta}{\cot 	heta - \cos 	heta}$Substitute the values $\cot 	heta = \frac{4}{3}, \sin 	heta = \frac{3}{5}, \cos 	heta = \frac{4}{5}$:$= \frac{\frac{4}{3} - \frac{3}{5}}{\frac{4}{3} - \frac{4}{5}} = \frac{\frac{20-9}{15}}{\frac{20-12}{15}} = \frac{11}{15} 	imes \frac{15}{8} = \frac{11}{8}$.

If $\theta$ is in the first quadrant and $\tan \theta = \frac{3}{4},$ then $\frac{\tan \left(\frac{\pi}{2}-\theta\right)-\sin (\pi-\theta)}{\sin \left(\frac{3 \pi}{2}+\theta\right)-\cot (2 \pi-\theta)} = $

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