If tan A = m/(m−1) and tan B = 1/(2m−1), Prove that A − B = π/4

Question

If \[ \tan A=\frac{m}{m-1} \] and \[ \tan B=\frac{1}{2m-1} \] prove that:

\[ A-B=\frac{\pi}{4} \]

Proof

Using the identity:

\[ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan B} \]

Substituting the given values:

\[ \tan(A-B) = \frac{\frac{m}{m-1}-\frac{1}{2m-1}} {1+\left(\frac{m}{m-1}\times\frac{1}{2m-1}\right)} \]

Taking LCM in the numerator:

\[ = \frac{\frac{m(2m-1)-(m-1)}{(m-1)(2m-1)}} {\frac{(m-1)(2m-1)+m}{(m-1)(2m-1)}} \]

Simplifying the numerator:

\[ m(2m-1)-(m-1) = 2m^2-m-m+1 \]

\[ = 2m^2-2m+1 \]

Simplifying the denominator:

\[ (m-1)(2m-1)+m \]

\[ = 2m^2-m-2m+1+m \]

\[ = 2m^2-2m+1 \]

Therefore,

\[ \tan(A-B) = \frac{2m^2-2m+1} {2m^2-2m+1} \]

\[ =1 \]

Hence,

\[ \tan(A-B)=1 \]

We know that:

\[ \tan\frac{\pi}{4}=1 \]

Therefore,

\[ A-B=\frac{\pi}{4} \]

Hence proved.