In a ∆ABC is 5cosC+6cosB=4 and 6cosA+4cosC=5, then tanA2tanB2=
15
25
35
45
Explanation for the correct option.
Finding the tanA2tanB2:
Given,
5cosC+6cosB=4 and 6cosA+4cosC=5.
These are in the form of bcosC+ccosA+a,ccosA+acosC=b.
On comparing these equation we get,
a=4,b=5,c=6
s=a+b+c2=152
tanA2tanB2=s-b(s-c)s(s-a)×(s-a)(s-c)s(s-b)=(s-c)2s2=(s-c)s=152-6152=315=15
Hence, the correct option is A.