If $ heta$ lies in the first quadrant and $5 an heta=4$, then $\frac{5 \sin heta-3 \cos heta}{\sin heta+2 \cos heta}$ is equal to

Question

If $	heta$ lies in the first quadrant and $5 	an 	heta=4$, then $\frac{5 \sin 	heta-3 \cos 	heta}{\sin 	heta+2 \cos 	heta}$ is equal to

Aicharya · Accepted Answer

Given, $	heta$ lies in the first quadrant $\begin{array}{rlrl}	ext { and } & 5 	an 	heta & =4 \ \Rightarrow & 	an 	heta & =\frac{4}{5} \ & 	herefore & \sin 	heta=\frac{4}{\sqrt{41}}, \cos 	heta & =\frac{5}{\sqrt{41}}\end{array}$ Now, $\begin{aligned} \frac{5 \sin 	heta-3 \cos 	heta}{\sin 	heta+2 \cos 	heta} & =\frac{5 	imes \frac{4}{\sqrt{41}}-\frac{3 	imes 5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}+2 	imes \frac{5}{\sqrt{41}}} \ & =\frac{20-15}{4+10} \ & =\frac{5}{14}\end{aligned}$

If $\theta$ lies in the first quadrant and $5 \tan \theta=4$, then $\frac{5 \sin \theta-3 \cos \theta}{\sin…

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