(Prove that):
(cos 7A + cos 3A - Cos 5A - COS A) / (sin 7A - sin 5A - sin3A + sinA) = cot 2A

Step-by-step explanation:

(cos7A + cos3A - cos5A - cosA)÷(sin7A - sin3A - sin5a + sinA)

By using transformations,

[(cos7A - cos5A) + (cos3A - cosA)]÷[(sin7A - sin5A) - (sin3A - sinA)]

we get...

=[(2sin6A.sinA)+(2sin2A.sinA)]÷[(2cos6A.sinA)-(2cos2A.sinA)]...

=[{2sinA(sin6A+sin2A)}]÷[{2sinA(cos6A-cos2A)]

=[{2sin4A.cos2A}÷{2sin4A.sin2A}]

=[(cos2A)÷(sin2A)]

=cot2A

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c