ΔABC is right angled at B, if cot A = \(\frac{1}{2}\) , find the value of \(\frac{sinA\;(cos C)}{cosC\;(sin C)}\).

2

cotA = \(\frac{1}{2}\)  (where 1 → P and 2 → B)

H = \(\sqrt {2^2 + 1}\) = \(\sqrt {5}\)

Put in \(\frac{sinA\;(cos C)}{cosC\;(sin C)}\)

⇒ \(\frac{\frac{2}{\sqrt {5}} × \frac{2}{\sqrt {5}}}{\frac{2}{\sqrt {5}} × \frac{1}{\sqrt {5}}}\) 

= \(\frac{4}{5}\) × \(\frac{5}{2}\) = 2