Question

If (Root 3)sinA = cos A. Then find the value of {tan A.sin A.(1 + cot A)}/ {sin A + cot A}.

Answer 1

$$\begin{gathered} \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}, \\ \sqrt{3}\mathrm{sinA}=\mathrm{cosA} \\ \mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{with}\sqrt{3}\mathrm{cosA}\mathrm{to}\mathrm{get}, \\ \frac{\mathrm{sinA}}{\mathrm{cosA}}=\frac{1}{\sqrt{3}} \\ \mathrm{tanA}=\frac{1}{\sqrt{3}} \\ \mathrm{tanA}=\tan 30° \\ \mathrm{A}=30° \\ \mathrm{Thus}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{given}\mathrm{expresson}\mathrm{is}, \\ \frac{\mathrm{tanA}\mathrm{sinA}\left(1+\mathrm{cotA}\right)}{\mathrm{sinA}+\mathrm{cotA}}=\frac{\tan 30°\sin 30°\left(1+\cot 30°\right)}{\sin 30°+\cot 30°} \\ =\frac{{\displaystyle \frac{1}{\sqrt{3}}}{\displaystyle \frac{1}{2}}\left(1+\sqrt{3}\right)}{{\displaystyle \frac{1}{2}}+\sqrt{3}} \\ =\frac{{\displaystyle \left(1+\sqrt{3}\right)\frac{1}{2\sqrt{3}}}}{{\displaystyle \frac{{\displaystyle 1+2\sqrt{3}}}{2}}} \\ =\frac{{\displaystyle \left(1+\sqrt{3}\right)}}{{\displaystyle \sqrt{3}\left(1+2\sqrt{3}\right)}} \\ =\frac{{\displaystyle \left(1+\sqrt{3}\right)}}{{\displaystyle \left(\sqrt{3}+2\times 3\right)}} \\ =\frac{{\displaystyle \left(1+\sqrt{3}\right)}}{{\displaystyle \left(6+\sqrt{3}\right)}} \\ \mathrm{Rationalize}\mathrm{the}\mathrm{denominator}\mathrm{to}\mathrm{get}, \\ \frac{\mathrm{tanA}\mathrm{sinA}\left(1+\mathrm{cotA}\right)}{\mathrm{sinA}+\mathrm{cotA}}=\frac{{\displaystyle \left(1+\sqrt{3}\right)\left(6-\sqrt{3}\right)}}{{\displaystyle \left(6+\sqrt{3}\right)\left(6-\sqrt{3}\right)}} \\ =\frac{{\displaystyle 6-\sqrt{3}+6\sqrt{3}-3}}{{\displaystyle {\left(6\right)}^2-{\left(\sqrt{3}\right)}^2}} \\ =\frac{{\displaystyle 3+5\sqrt{3}}}{{\displaystyle 36-3}} \\ =\frac{{\displaystyle 3+5\sqrt{3}}}{{\displaystyle 33}} \end{gathered}$$