Question

If sin^2 a + sin^2 b + sin^2 c + sin^2 d = 0 , then minimum of osa + Cosb + Cosc + Cosd is (A) 4 (C) - 6 (D) - 4 (B) 0​

Answer 1

Step-by-step explanation:

Consider the problem,

sin

2

A+sin

2

B+sin

2

c=2+2cosA.cosB.cosc

We can write sin

2

A as,

sin

2

A=

2

1−cos(2A)

Therefore,

LHS=

2

1−cos(2A)

+

2

1−cos(2B)

+

2

1−cos(2C)

=

2

3

−(cos(2A)+cos(2B)+cos(2C))

=

2

1

(3−(2cos(A+B)cos(A−B)+cos(2C)))

C=180−(A+B)

cos(C)=cos(180−(A+B))

cos(C)=−cos(A+B)

Therefore,

=

2

3

−(−2cos(C)cos(A−B)+cos(2C))

cos(2C)=2cos

2

(C)−1

And,

=

2

1

(3−(−2cosC)cos(A−B)+2cos

2

(C)−1)

=

2

1

(4−(2cos(C)cos(C)−cos(A−B)))

=

2

1

(4−2cos(C)(−cos(A+B)−cos(A−B)))

=

2

1

(4+2cos(C)(cos(A+B)cos(A−B)))

=

2

1

(4+2cos(C)×2cos(A)cos(B))

=2+2cos(A)cos(B)cos(C)

Therefore, If A+B+C=180, sin

2

A+sin

2

B+sin

2

c=2+2cosA.cosB.cosc

Answer 2

Answer:

A

Step-by-step explanation:

You just need to solve it well ,use

sin2=1-Cos2.

using this identity you must get 4