Question

If $\cos ecx+cotx=\frac{11}{2}$, then tan x =
(a) $\frac{21}{22}$

(b) $\frac{15}{16}$

(c) $\frac{44}{117}$

(d) $\frac{117}{43}$

Answer 1

(c) $\frac{44}{117}$

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \mathrm{cosec}x+\cot x=\frac{11}{2}\left(1\right) \\ \Rightarrow \frac{1}{\mathrm{cosec}x+\cot x}=\frac{2}{11} \\ \Rightarrow \frac{{\mathrm{cosec}}^{2}x-{\cot }^{2}x}{\mathrm{cosec}x+\cot x}=\frac{2}{11} \\ \Rightarrow \frac{\left(\mathrm{cosec}x+\cot x\right)\left(\mathrm{cosec}x-\cot x\right)}{\left(\mathrm{cosec}x+\cot x\right)}=\frac{2}{11} \\ \therefore \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}x-\mathrm{c}\mathrm{o}\mathrm{t}x=\frac{2}{11}\left(2\right) \\ \mathrm{Subtracting}\left(2\right)\mathrm{from}\left(1\right): \\ 2\mathrm{c}\mathrm{o}\mathrm{t}x=\frac{11}{2}-\frac{2}{11} \\ \Rightarrow 2\mathrm{c}\mathrm{o}\mathrm{t}x=\frac{121-4}{22} \\ \Rightarrow 2\mathrm{c}\mathrm{o}\mathrm{t}x=\frac{117}{22} \\ \Rightarrow \mathrm{c}\mathrm{o}\mathrm{t}x=\frac{117}{44} \\ \Rightarrow \frac{1}{\tan x}=\frac{117}{44} \\ \Rightarrow \tan x=\frac{44}{117} \end{gathered}$$