If $${\left( {r\cos \theta - \sqrt 3 } \right)^2}$$ + $${\left( {r\sin \theta - 1} \right)^2}$$ = 0, then the value of $$\frac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan \theta }}$$ is equal to?

Question

নিচের অপশন গুলা দেখুন

• $$\frac{4}{5}$$
• $$\frac{{\sqrt 3 }}{4}$$
• $$\frac{{\sqrt 5 }}{4}$$
• $$\frac{3}{5}$$

Answer 1

সঠিক উত্তর হচ্ছে: $$\frac{4}{5}$$

ব্যাখ্যা:

$$\eqalign{ & {\left( {r\cos \theta - \sqrt 3 } \right)^2} + {\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {r\cos \theta - \sqrt 3 } \right)^2} = 0,{\text{ }}{\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow r{\text{ cos}}\theta = \sqrt 3 ........(i) \cr & \Rightarrow r\sin \theta = 1.........(ii) \cr & {\text{Squaring and adding equation (i) and (ii)}} \cr & \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 3 + 1 \cr & \Rightarrow {r^2}\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {r^2} = 4 \cr & \Rightarrow r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }}{\text{ and }}r{\text{ cos}}\theta = \sqrt 3 \cr & \cos \theta = \frac{{\sqrt 3 }}{r}, \cr & \sec \theta = \frac{r}{{\sqrt 3 }} \cr & \frac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan \theta }} \cr & = \frac{{\frac{r}{{\sqrt 3 }} + \frac{r}{{\sqrt 3 }}}}{{\frac{{{r^2}}}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{{r\left( {\frac{2}{{\sqrt 3 }}} \right)}}{{\frac{{{r^2} + 1}}{{\sqrt 3 }}}} \cr & = \frac{{2r}}{{{r^2} + 1}} \cr & = \frac{{2 \times 2}}{{{2^2} + 1}} \cr & = \frac{4}{5} \cr & \cr & {\bf{Alternate:}} \cr & r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }} \cr & \theta = {30^ \circ } \cr & = \frac{{2\tan {{30}^ \circ } + \sec {{30}^ \circ }}}{{2\sec {{30}^ \circ } + \tan {{30}^ \circ }}} \cr & = \frac{{2 \times \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }}}}{{2 \times \frac{2}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{4}{5} \cr} $$