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Let's consider the given figure:
Photo by Math Principles in Everyday Life |
Solution:
First, from the x-axis and y-axis, let's draw line 1 and line 2 that will pass at the origin and let's assign point A there. Let ϕ is the measurement of an angle from line 1 to line 2 while θ is the measurement of an angle from line 2 to x-axis. Next, assign point B at line 1. From point B, draw a vertical line that is perpendicular to x-axis and let's assign point E at the x-axis. Consider ΔABE,
In this derivation, we have to eliminate the line segments and we need the angles only. So, we have to do further derivations. From point B, draw a line that is perpendicular to line 2 and let's assign point C at line 2. From point C, draw a vertical line that is perpendicular to the x-axis and let's assign point D at the x-axis. From point C also, draw a horizontal line which is perpendicular to y-axis and let's assign point F at line EB. Consider ΔACD,
or
or
(Reason: If line FC is parallel to line ED and line FE is parallel to line CD, then the line segment FC is congruent to line segment ED and the line segment EF is congruent to line segment CD. The resulting figure EFCD is a parallelogram. Since each four lines are perpendicular to each other, then the parallelogram is a rectangle.)
Let's have another one,
using the same figure, we have
but
(Reason: If line FC is parallel to line ED and line FE is parallel to line CD, then the line segment FC is congruent to line segment ED and the line segment EF is congruent to line segment CD. The resulting figure EFCD is a parallelogram. Since each four lines are perpendicular to each other, then the parallelogram is a rectangle.)
If you will substitute -ϕ to the above equation, then
How about for Tan (θ + ϕ)?
Ok, let's do for Tan (θ + ϕ). We don't have to use the figure again to derive Tan (θ + ϕ). We know that
If we divide both the numerator and denominator by cos ϕ cos θ, then the above equation will be
If you will substitute -ϕ to the above equation, then