I don’t really know let me ask my friends 1
Notice that
11/12 = 1/6 + 3/4
so that
tan(11π/12) = tan(π/6 + 3π/4)
Then recalling that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
⇒ tan(x + y) = (tan(x) + tan(y))/(1 - tan(x) tan(y))
it follows that
tan(11π/12) = (tan(π/6) + tan(3π/4))/(1 - tan(π/6) tan(3π/4))
tan(11π/12) = (1/√3 - 1)/(1 + 1/√3)
tan(11π/12) = (1 - √3)/(√3 + 1)
tan(11π/12) = - (√3 - 1)²/((√3 + 1) (√3 - 1))
tan(11π/12) = - (4 - 2√3)/2
tan(11π/12) = - (2 - √3) … … … [A]
Answer:
Step-by-step explanation:
3y+7=2x
3y=2x-7
y=(2/3)x-7/3
when parallel, the y=ax+b, the a keeps the same
so it‘s y=(2/3)x-a
and it pass (2,6), so 6=(2/3)*2 -a
6=4/3-a
-a=(18-4)/3
-a=14/3
a=-14/3
