$\cos (x+iy)\cos (u+iv) = 1$ $\cos (x+iy)=\sec (u+iv)$ $\sin (x+iy)= \sqrt{1-\cos^2 (x+iy)}= \sqrt{1-\sec^{2}(u+iv)}= \sqrt{-\tan^2 (v+iv)}$ $\therefore , \ \sin (x+iy) = i \tan (u+iv)$ $Now, \ \tan (x+iy)= \dfrac {\sin (x+iy)}{\cos (x+iy)} = \dfrac {i \tan (u+iv)}{\sec (u+iv)} = i \sin (u+iv) $ $So, \tan (x-iy)=-\sin (u-iv)$ $\tan 2iy=\tan [ (x+iy)- (x-iy)]$ $= \dfrac {\tan (x+iy)- \tan (x-iy)}{1+\tan (x-iy)\tan (x+iy)}= \dfrac{i \sin (u+iv)+ i \sin (u-iv)}{1-i^2 \sin (u+iv)\sin (u-iv)}$ $\tan 2iy = \dfrac {i 2 \sin u \cos iv}{1 +\frac {1}{2} \left [ \cos 2 iv - \cos 2 u \right ]}$ $i \tan 2y = \dfrac {2i \sin u \cosh v}{1+\frac {1}{2} \left [ \cosh 2v-\cos 2u \right ]}$ $\tanh 2y= \dfrac {2\sin u \cosh v}{1+ \frac {2 \cosh^2 v-1-1+2 \sin^2 u}{2}} = \dfrac {2 \sin u \cosh v}{\cosh^2 v + \sin^2 u}$ Dividing Numerator and Denominator by cosh2 v, $\tanh 2y=\dfrac {\frac {2\sin u}{\cosh v}}{1+ \frac {\sin^2 u}{\cosh v}}$ $\dfrac {2 \tanh y}{1+\tanh^2 y} = \dfrac {2 \left ( \frac {\sin u}{\cosh v} \right )}{1+ \left ( \frac {\sin^2 u}{\cosh^2 v} \right )}$ Comparing both sides, $\tanh y=\dfrac {\sin u}{\cosh v}$ $\therefore, \ \tanh^2 y \cosh^2 v= \sin ^2 u$