• Consider LHS :$e^{2 \ ai \cot ^{-1}b} \left [ \dfrac {bi-1}{bi+1} \right ]^{-a} $....................1

$\dfrac{bi-1}{bi+1}=\dfrac{i\bigg(b-\dfrac{1}{i}\bigg)}{i\bigg(b+\dfrac{1}{i}\bigg)}=\dfrac{b+i}{b-i}$

• Let polar form of complex no :b+i=r(cos$\theta$+isin$\theta$)=rei$\theta$

where r=$\sqrt{b^2+1}$ and $\theta$=$\tan^{-1}\bigg(\dfrac{1}{b}\bigg)$

• Therefore$b-i=r(\cos\theta-i\sin\theta)=re^{-i\theta}$

Hence $\dfrac{bi-1}{bi+1}$=$\dfrac{re^{i\theta}}{re^{-i\theta}}$=$e^{i2\theta}$=$e^{i2tan^-1(\frac{1}{b})}$=$e^{i2cot^-1(b)}$

• From 1 we have$e^{2 \ ai \cot ^{-1}b} *(e^{i2cot^{-1}b})^{-a}$=$e^{2aicot^{-1}b-2aicot^{-1}b}$=$e^{0}=1$