Answer:
[tex]\huge\boxed{\cot\theta=\dfrac{1}{xy}}[/tex]
Step-by-step explanation:
[tex]\bold{METHOD\ 1}[/tex]
[tex]\sin\theta=x\\\\\sec\theta=y\\\\\cot\theta=?\\\\\text{We know:}\\\\\sec x=\dfrac{1}{\cos x};\ \cot x=\dfrac{\cos x}{\sin x}\\\\\sec\theta=y\to\dfrac{1}{\cos \theta}=y\to\dfrac{\cos\theta}{1}=\dfrac{1}{y}\to\cos\theta=\dfrac{1}{y}\\\\\cot \theta=\dfrac{\frac{1}{y}}{x}=\dfrac{1}{xy}[/tex]
[tex]\bold{METHOD\ 2}[/tex]
[tex]\text{We know}\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\cot x=\dfrac{\cos x}{\sin x}=\dfrac{1}{\tan x}\\\\\sec x=\dfrac{1}{\cos x}\\\\\text{therefore}\\\\(sin x)(\sec x)=(\sin x)\left(\dfrac{1}{\cos x}\right)=\dfrac{\sin x}{\cos x}=\tan x\\\\\dfrac{1}{(\sin x)(\sec x)}=\dfrac{1}{\tan x}=\cot x[/tex]
[tex]\\\sin \theta=x;\ \sec\theta=y\\\\\text{substitute}\\\\\cot\theta=\dfrac{1}{xy}[/tex]
