Question

tan[π\4+Φ]•tan[3π\4+Φ]​

Answer 1

Answer:

Convert radians to degrees:

tan(45+θ)*tan(135+θ)=-1

Now, do you notice something?

Assuming you can treat gradients as tan ratios, it is exactly the same as the formula for two perpendicular gradients! (m1*m2=-1, and the actual proof of this involves similar triangles: get a constructed right angled triangle to have its hypotenuse as its new base, an get a perpendicular altitude. The altitude and the two non-90 angles in the triangle give enough information that it can be split into two smaller, but similar triangles (AA, third angle to be 90 through deduction). Judging by the position of the angles in the smaller triangles, one of the smaller triangles is essentially the other, but rotated 90 degrees and scaled up/down. That means the opposites and the adjacents have swapped places, if you wanted to transform smaller triangle A to smaller triangle B and if the two non-90 degree angles (of the large triangle) are the angles of reference. The tangent of one (triangle, we’re focusing on triangles, not gradients) therefore is the reciprocal of the other. One line, however, is going in the opposite direction to the other, so one of the gradients must be negative. We know that the negative reciprocal of a number multiplied by the original number is -1. Just substitute the gradients, and you’ll see.)

Say we have a triangle ABC on a Cartesian plane. tan(45+θ) is the gradient of line AB. tan(135+θ) is the gradient of BC. A and C are on the x axis, and B hangs above them, so that the area of the triangle divided by half of AC is the most direct distance B is to the x axis. The supplementary angle of 135+θ is 45-θ, and the reason we need to know this is because a triangle is a closed shape. 45-θ+(45+θ)=90, and the third angle must be 90 degrees.

Because we know that AB is perpendicular to BC, we can also conclude that the tan ratios multiplied must equal -1.

Answer 2

$\tan = ( \frac{ \pi}{4} + a) \times \tan( \frac{3 \pi + A}{4} )$

$using \: tangent \: sum \: of \: angle \: formula \ratio -$

$tan(A+B) = \frac{ \tan A+ \tan B}{1 - \tan A \times \tan B }$

$in \: conjuction \: with \ratio$

$\tan( \frac{ \tan \pi }{4} ) = 1 \: and \: \tan( \frac{ \tan 3\pi }{4} ) = - 1$

$now$

$\tan( \frac{\pi}{4} + A) \times \tan( \frac{3 \pi }{4} + A)$

$= \frac{ \tan (\frac{ \pi }{4} ) + \tan A}{1 - \tan( \frac{ \pi }{4}) \times \tan A }$

$= > \frac{ \tan (\frac{3 \pi }{4} ) + \tan A}{1 - \tan( \frac{ 3\pi }{4}) \times \tan A }$

$= \frac{1 + \tan A }{1 - \tan A} \times \frac{ - 1 + \tan A }{1 + \tan A}$

$= \frac{ - 1 + \tan A }{1 - \tan A} \times \frac{ 1 - \tan A }{1 + \tan A}$

$\boxed{= - 1}$