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Question

# If 4tanθ = 3, evaluate $\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta +1}{4\mathrm{sin}\theta +\mathrm{cos}\theta -1}$

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Solution

## Given: 4tanθ = 3 ⇒ tan θ = $\frac{3}{4}$ Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3k and AB = 4k, where k is a positive number. By Pythagoras theorem, we get ${\mathrm{AC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}={\left(3k\right)}^{2}+{\left(4k\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}=9{k}^{2}+16{k}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{25{k}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=±5k\phantom{\rule{0ex}{0ex}}$ Ignoring AC = − 5k , as k is a positive number, we get AC = 5k If $\mathrm{tan}\theta =\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$, then $\mathrm{sin}\theta =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{5}$ and $\mathrm{cos}\theta =\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$ Putting the values in $\left(\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta +1}{4\mathrm{sin}\theta +\mathrm{cos}\theta -1}\right)$, we get $\left(\frac{4×\frac{3}{5}-\frac{4}{5}+1}{4×\frac{3}{5}+\frac{4}{5}-1}\right)=\left(\frac{\frac{12-4+5}{5}}{\frac{12+4-5}{5}}\right)=\frac{13}{11}$

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