Answer:[tex]-\frac{189}{208}[/tex]
Step-by-step explanation:
Given
[tex]\sin x=\frac{8}{10}[/tex]
x lies in the second quadrant where sin and cosec is positive and rest trigonometric functions are negative
so, [tex]\cos x=-\sqrt{1-sin ^2 x}\\\cos x=-\sqrt{1-(\frac{8}{10})^2}=-\sqrt{\frac{36}{100}}\\\cos x=-\sqrt{\frac{6}{10}}[/tex]
[tex]\tan x=\frac{\sin x}{\cos x}=-\frac{4}{3}[/tex]
similarly, [tex]\tan y=-\frac{12}{5}[/tex]
Also, y lies in 4 th Quadrant where cos and sec is positive and rest trigonometric functions are negative
[tex]\sec y=\sqrt{1+\tan ^2y}=\sqrt{1+(-\frac{12}{5})^2}\\\sec y=\frac{13}{5}\\\cos y=\frac{5}{13}\\\sin y =\frac{\tan y}{\cos y}=-\frac{12}{13}[/tex]
Now putting the values
=[tex]\dfrac{\cos x \cos y+\sin x\sin y}{\tan x-\tan y}=\dfrac{(\frac{-6}{10})(\frac{5}{13})+(\frac{8}{10})(\frac{-12}{13})}{(\frac{-8}{6})-(\frac{-12}{5})}\\\dfrac{\cos x \cos y+\sin x\sin y}{\tan x-\tan y}=\dfrac{\frac{-126}{130}}{\frac{16}{15}}=-\frac{189}{208}[/tex]
