‌sech−1x=log‌2 ‌‌ and ‌cosech−1y=−log‌3sech−1x=log‌2 ‌⇒‌‌log(‌
1+√1−x2
x
)=log‌2 ‌⇒‌‌‌
1+√1−x2
x
=2 ‌⇒‌‌1+√1−x2=2x ‌√1−x2=2x−1 Squaring both sides, we get ‌⇒‌‌1−x2=4x2+1−4x ‌⇒‌‌5x2−4x=0 ‌⇒‌‌x(5x−4)=0 ‌‌‌x=‌
4
5
‌∵cosech−1y=−log‌3 ‌⇒‌‌log(‌
1+√1+y2
y
)=log‌
1
3
‌⇒‌‌‌
1+√1+y2
y
=‌
1
3
‌⇒‌‌1+√1+y2=‌
y
3
‌⇒‌‌√1+y2=‌
y
3
−1 Squaring both sides, we get ‌⇒‌‌1+y2=‌
