Question

30.1f 4+ +C = 270". then
cosA + cos2B + cos2C+4sinAsinBsinC
(a) 0
(c) 2
(b) 1
(d)3​

Answer 1

Answer:

Mam first of all Question is wrongly written , it should be :-

if A+B+C = 270° ,

find Cos2A+cos2B+cos2C + 4sinAsinBsinC

Now see solution :-----

(A+B) = (270-C)

Cos2A + Cos2B = 2Cos(A+B)Cos(A-B)

= 2Cos(270-C)Cos(A-B)

= -2Sin(C)Cos(A-B)----(1)

Cos2C = 1-2sin²C ----------------(2)

Now,

4SinASinBSinC = 2(2SinASinB)SinC

= 2[Cos(A-B) - Cos(A+B)]SinC

= 2[Cos(A-B) - Cos(270-C)]SinC

= 2[cos(A-B) + SinC]SinC

= 2Cos(A-B)SinC + 2Sin²C --------(3)

Adding all 3 equations now,

we get :-

$\boxed{Answer}$ = 1

$\color{red}{Mark\: as\: Brainlist}$