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Question

# If $\mathrm{cos}ecA+cotA=\frac{11}{2}$, then tan A = (a) $\frac{21}{22}$ (b) $\frac{15}{16}$ (c) $\frac{44}{117}$ (d) $\frac{117}{43}$

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Solution

## (c) $\frac{44}{117}$ $\mathrm{We}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\mathrm{cosec}\mathrm{A}+\mathrm{cot}\mathrm{A}=\frac{11}{2}\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cosec}\mathrm{A}+\mathrm{cot}\mathrm{A}}=\frac{2}{11}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{cosec}}^{2}\mathrm{A}-{\mathrm{cot}}^{2}\mathrm{A}}{\mathrm{cosec}\mathrm{A}+\mathrm{cot}\mathrm{A}}=\frac{2}{11}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(\mathrm{cosec}\mathrm{A}+\mathrm{cot}\mathrm{A}\right)\left(\mathrm{cosec}\mathrm{A}-\mathrm{cot}\mathrm{A}\right)}{\left(\mathrm{cosec}\mathrm{A}+\mathrm{cot}\mathrm{A}\right)}=\frac{2}{11}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{A}-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{A}=\frac{2}{11}\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(2\right)\mathrm{from}\left(1\right):\phantom{\rule{0ex}{0ex}}2\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{A}=\frac{11}{2}-\frac{2}{11}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{A}=\frac{121-4}{22}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{A}=\frac{117}{22}\phantom{\rule{0ex}{0ex}}⇒\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{A}=\frac{117}{44}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{tan}\mathrm{A}}=\frac{117}{44}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\mathrm{A}=\frac{44}{117}$

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