Question

Find the value of `ln(coshx+sinhx)+ln(coshx+sinhx)`?

Answer 1

`ln(coshx+sinhx)+ln(coshx-sinhx)=0`

Explanation:

`coshx=(e^x+e^(-x))/2` and `sinhx=(e^x-e^(-x))/2`

Hence `coshx+sinhx=e^x` and `ln(coshx+sinhx)=x`

Similarly `coshx-sinhx=e^(-x)` and `ln(coshx-sinhx)=-x`

Hence adding the two, we get

`ln(coshx+sinhx)+ln(coshx-sinhx)=0`