If $2 an A=3$, then the value of $\frac{4 \sin A+3 \cos A}{4 \sin A-3 \cos A}$ is\

Question

If $2 	an A=3$, then the value of $\frac{4 \sin A+3 \cos A}{4 \sin A-3 \cos A}$ is\

Professor PC Thomas Classes · Answer

(c): We have, $2 	an A=3$ $$ \Rightarrow \quad 	an A=\frac{3}{2}=\frac{P}{B} $$ Let $P=3 k$ and $B=2 k$ $$ H=\sqrt{(2 k)^{2}+(3 k)^{2}} $$ (By Pythagoras theorem)\ $\Rightarrow \quad H=\sqrt{13} k$\ $	herefore \quad \sin A=\frac{P}{H}=\frac{3}{\sqrt{13}}, \cos A=\frac{B}{H}=\frac{2}{\sqrt{13}}$\ Now, $\frac{4 \sin A+3 \cos A}{4 \sin A-3 \cos A}=\frac{4\left(\frac{3}{\sqrt{13}}ight)+3\left(\frac{2}{\sqrt{13}}ight)}{4\left(\frac{3}{\sqrt{13}}ight)-3\left(\frac{2}{\sqrt{13}}ight)}=3$\