Question

In $\Delta ABC,\angle C={90}^{\circ },\cot A=\sqrt{3}and\cot B=\frac{1}{\sqrt{3}},$ then prove : that $\sin A\cos B+\cos A,\sin B=1$

Answer 1

Here we are given that
$\cot A=\frac{\sqrt{3}}{1}=\frac{b}{p}$
Here , $b=\sqrt{3}k,p=1k$
$\therefore$ By using Baudhayana theorem
$A{B}^2=A{C}^2+B{C}^2$
$=(\sqrt{3}k{)}^2+(k{)}^2=3k^2+k^2$
$=4k^2$
$\therefore$ $AB=2k$
$\therefore$ $\sin A=\frac{BC}{AB}=\frac{1k}{2k}=\frac{1}{2}$
$\cos A=\frac{AC}{AB}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$
Similarly, by using $\cot B=\frac{1}{\sqrt{3}}=\frac{BC}{AC}$
$\Rightarrow$ $BC=1,AC=\sqrt{3}k$
$\therefore$ $A{B}^2=B{C}^2+A{C}^2$
$=(1k{)}^2+(\sqrt{3}k{)}^2$
$=1k^2+3k^2=4k^2$
$\Rightarrow$ $AB=2k$
$\cos B=\frac{BC}{AB}=\frac{1k}{2k}=\frac{1}{2}$
$\sin B=\frac{AC}{AB}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$
$\therefore$ L.H.S $=\sin A.\cos B+\cos A.\sin B$
$=\frac{1}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}$
$=1$ = R.H.S
Hence Proved.