If $heta$ be acute angle and $cos heta=frac{15}{17}$, then the value of $cot left(90^{circ}- heta ight)$ is

$cos heta=frac{15}{17} Rightarrow sec heta=frac{1}{cos heta}=frac{17}{15}$ $herefore cot left(90^{circ}- heta ight)= an heta=sqrt{sec ^{2} heta-1}$ $=sqrt{left(frac{17}{15} ight)^{2}-1}=sqrt{frac{289}{225}-1}$ $=sqrt{frac{289-225}{225}}=sqrt{frac{64}{225}}=frac{8}{15}$ Alternative:

Given that $cos heta=frac{15}{17}=frac{b}{h}$

Height $=sqrt{( ext { Hypotenuse })^{2}-( ext { Base })^{2}}$

$=sqrt{(17)^{2}-(15)^{2}}=sqrt{64}=8$

$cot left(90^{circ}- heta ight)= an heta=frac{ ext { Height }}{ ext { Base }}$

$=frac{8}{15}$