If A and B are complementary acute angles, what is the value of (Tan A + Tan B)/(Csc(360-2A))?
This value is -2 regardless of A and B.
Two complementary acute angles by definition add up to `90^@ , ` i.e. `A + B = 90^@ . ` A place where one can observe two complementary acute angles is a right triangle. Its acute angles plus the right angle `90^@ ` have the sum of measures of `180^@ , ` which is the cause.
In a right triangle, the tangent of an acute angle is the ratio between the opposing leg of the angle and the adjacent leg. Because of this, the tangents of `A ` and `B ` are reverse numbers, `tan B = 1 / tan A .`
Also we know that
`csc ( 360^@ - 2 A ) = 1 / sin ( 360^@ - 2 A ) = 1 / sin ( - 2 A ) = -1 / sin ( 2 A ) . `
The second equality is true because sine is `360^@ `-periodic.
Now combine everything we know:
`( tan A + tan B ) / csc ( 360^@ - 2A ) = ( sin A / cos A + cos A / sin A ) / ( -1 / sin ( 2 A ) ) =` ` ( ( sin^2 A + cos^2 A ) / ( sin A cos A ) ) / ( -1 / ( 2 sin A cos A ) ) = -2 ,`
which is the answer.