# if cosec theta=5/3, evaluate 4sec theta-2 tan theta + 5 sin theta/ 20 cos theta - 3 cosec theta + 9 cot theta

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$\mathrm{cos}ec\theta =\frac{5}{3}=\frac{H}{P}\phantom{\rule{0ex}{0ex}}B=\sqrt{{H}^{2}-{P}^{2}}=\sqrt{{5}^{2}-{3}^{2}}=4\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}\mathrm{sin}\theta =\frac{3}{5}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{4}{5}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{3}{4}\phantom{\rule{0ex}{0ex}}cot\theta =\frac{4}{3}\phantom{\rule{0ex}{0ex}}sec\theta =\frac{5}{4}\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}\frac{4sec\theta -2\mathrm{tan}\theta +5\mathrm{sin}\theta }{20\mathrm{cos}\theta -3\mathrm{cos}ec\theta +9cot\theta }\phantom{\rule{0ex}{0ex}}=\frac{\left(4×\frac{5}{4}\right)-\left(2×\frac{3}{4}\right)+\left(5×\frac{3}{5}\right)}{\left(20×\frac{4}{5}\right)-\left(3×\frac{5}{3}\right)+\left(9×\frac{4}{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{5-\frac{3}{2}+3}{16-5+12}\phantom{\rule{0ex}{0ex}}=\frac{\frac{13}{2}}{23}\phantom{\rule{0ex}{0ex}}=\frac{13}{46}$

• 10
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