Question

If 4 tan theta = 3 , then find (4 sin theta - cos theta-1) / (4 sin theta + cos theta+1)

Answer 1

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Answer 2

Answer:

$Value \: of \: \frac{4sin\theta-cos\theta-1}{4sin\theta+cos\theta+1}=\frac{1}{17}$

Step-by-step explanation:

$Given \: 4tan\theta = 3\\\implies tan\theta = \frac{3}{4}\:---(1)$

$sec\theta \\= \sqrt{1+tan^{2}\theta}\\=\sqrt{1+\left(\frac{3}{4}\right)^{2}}\\=\sqrt{1+\frac{9}{16}}\\=\sqrt{\frac{16+9}{16}}\\=\sqrt{\frac{25}{9}}\\=\frac{5}{3}\:---(2)$

$Value \: of \: \frac{4sin\theta-cos\theta-1}{4sin\theta+cos\theta+1}$

$Divide\: numerator\: and\\ denominator\: by \:cos\theta,\\ we get$

$=\frac{4tan\theta-1-\frac{1}{cos\theta}}{4tan\theta+1+\frac{1}{cos\theta}}$

$=\frac{4tan\theta-1-sec\theta}{4tan\theta+1+sec\theta}$

$=\frac{4\times \frac{3}{4}-1-\frac{5}{3}}{4\times \frac{3}{4}+1+\frac{5}{3}}$

/* From (1) and (2) */

$=\frac{3-1-\frac{5}{3}}{3+1+\frac{5}{3}}\\=\frac{2-\frac{5}{3}}{4+\frac{5}{3}}$

$=\frac{\frac{6-5}{3}}{\frac{12+5}{3}}$

$=\frac{1}{17}$

Therefore,

$Value \: of \: \frac{4sin\theta-cos\theta-1}{4sin\theta+cos\theta+1}=\frac{1}{17}$

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