Prove that #(1+sinh2A+cosh2A)/(1-sinh2A-cosh2A)=-cothA#?

1 Answer
Jun 15, 2016

Answer:

See the solution below.

Explanation:

We will use the identities #(1+cosh2A)=2cosh^2A#, #(1-cosh2A)=-2sinh^2A# and #sinh2A=2sinhAcoshA#. Hence

#(1+sinh2A+cosh2A)/(1-sinh2A-cosh2A)#

= #(2cosh^2A+2sinhAcoshA)/(-2sinhAcoshA-2sinh^2A)#

= #(2coshA(coshA+sinhA))/(-2sinhA(coshA+sinhA))#

= #-(coshA)/(sinh2A)#

= #-cothA#