Answer:
Option B
Explanation:
We know that , coth−1x=12loge(x+1x−1),|x|>1
tanh−1x=12loge(1+x1−x)
and cosech−1x=loge(1−√1+x2x),x<0
∴ coth−1(3)=12loge(42)=loge√2
tanh−1(13)=12loge(1+1/31−1/3)=12loge(42)=loge√2
and cosech−1(−√3)=loge(1−√1+3−√3)
= loge(1−2−√3)=loge(1√3)
Hence, coth−1(3)+tanh−113−cosech−1(−√3)=
= loge√2+loge√2−loge(1√3)
=loge(√2.√2)+loge(√3)=loge(2√3)