In ∆ABC,(a+b+c)tanA2+tanB2 is equal to
2ccotC2
2acotA2
tanC2
None of these
Explanation for the correct option:
Solving the expression:
From sine rule, we know in ∆ABC,
tanA2=s-cs-bss-a and tanB2=s-as-css-b
Also, we know s=a+b+c2
∴(a+b+c)tanA2+tanB2=2s(s-b)(s-c)s(s-a)+(s-a)(s-c)s(s-b)=2ss-cs(s-b)(s-a)+(s-a)(s-b)=2s(s-c)s-b+s-a(s-a)(s-b)[∵2s=a+b+c]=2s(s-c)c(s-a)(s-b)[∴2s-b-a=c]=2cs(s-c)(s-a)(s-b)=2ccotC2[∵cotC2=s(s-c)(s-a)(s-b)]∆ABC,(a+b+c)tanA2+tanB2=2ccotC2
Hence, option(A) is the correct answer.
In a ΔABC, side b is equal to