© examsiri.com
Question : 25 of 160
Marks:
+1,
-0
Solution:
‌ We have, ‌log(9+3√2(2+√5)+4√5)
log(9+6√2+3√10+4√5)
log(9+3√10+6√2+4√5)
log(3+√10)(3+2√2)
=‌‌log(3+√10)+log(3+√8)
=‌‌log(3+√32+1)+log(3+√32−1)
=‌‌sin‌h−1‌3+cosh−13
© examsiri.com
Go to Question: