If cot A = k, then sin A is equal to : (presume that A is an acute angle)

$\frac{1}{\sqrt{1+k^2}}$ $\frac{k^2}{\sqrt{1+k^2}}$ $\frac{1}{k}$ $-\frac{1}{k}$

$\frac{1}{\sqrt{1+k^2}}$

cot A = k { cot A = \(\frac{B }{P}\) } By pythagoras theorem, P² + B² = H² 1² + k² = H² H = \(\sqrt { 1 + k² }\) Now, sinA = \(\frac{P }{H}\) = \(\frac{1 }{ \sqrt { 1 + k² }\}\)