Question

Prove that tan(π/4+A)tan(3π/4+A)=-1?

Answer 1

We seek tro show:

`tan(pi/4+A)tan((3pi)/4+A) -= -1`

Using the tangent sum of angle formula:

`tan(A+B) -= (tanA + tanB)/(1-tanAtanB)`

In conjunction with:

`tan(pi/4) = 1` and `tan((3pi)/4) = -1`

Then by considering the LHS of the given expression, we have:

`LHS = tan(pi/4+A)tan((3pi)/4+A)`

`\ \ \ \ \ \ \ \ = (tan(pi/4) + tanA)/(1-tan(pi/4)tanA) \ * \ (tan((3pi)/4) + tanA)/(1-tan((3pi)/4)tanA)`

`\ \ \ \ \ \ \ \ = (1 + tanA)/(1-tanA) \ * \ (-1 + tanA)/(1+tanA)`

`\ \ \ \ \ \ \ \ = - (1 + tanA)/(1-tanA) \ * \ (1 - tanA)/(1+tanA)`

`\ \ \ \ \ \ \ \ = - 1`

`\ \ \ \ \ \ \ \ -= RHS \ \ \ \ ` QED