In any ∆ABC,(a+b+c)(b+c-a)(c+a-b)(a+b-c)4b2c2 equals:
sin2B
cos2A
cos2B
sin2A
Explanation for the correct option.
Finding the value of the expression:
As per the Heron's Formula, in ∆ABC, a+b+c=2s. So,
(a+b+c)(b+c-a)(c+a-b)(a+b-c)4b2c2=(2s)(2s-2a)(2s-2b)(2s-2c)4b2c2=16ss-as-bs-c4b2c2=4ss-as-bs-cbc×bc=4∆2bc×bc[ByAreaof∆=ss-as-bs-c]=412bcsinA2bc×bc[ByAreaof∆=12bcsinA]=sin2A
Hence, option D is correct.
In any ΔABC, (a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2 is equal to