Prove that tan(π/4+A)tan(3π/4+A)=-1?

1 Answer
Steve M
Jun 12, 2018

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

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