Question

Please solve this question......as soon as possible

Answer 1

Answer:

There appears to be something missing from the statement of this problem as it does not seem to be true in general.

However, it is true if we restrict ourselves to the case that A, B, C are angles of a triangle.  So we should probably suppose that this is what was actually meant here.

No promises that the following is elegant. :-)

Make use of:

$2\sin^2 x = 1 - \cos 2x \\2\cos^2 x = 1 + \cos 2x \\4\sin^4 x = 1 - 2\cos 2x + \cos^2 2x \\A + B + C = \pi \Rightarrow \cos2C=\cos\bigl(2\pi-2(A+B)\bigr)=\cos(2A+2B)$

Then to show that the expression are equal, we'll show that their difference is equal to zero:

$2\times\bigl[(\sin^2 A +\sin^2 B + \sin^2 C)^2 - 2(\sin^4 A + \sin^4 B + \sin^4 C) - 4\sin^2 A\sin^2 B\sin^2 C\bigr] \\=(2\sin^2 A +2\sin^2 B + 2\sin^2 C)^2 - 2(4\sin^4 A + 4\sin^4 B + 4\sin^4 C) - 2(2\sin^2 A)(2\sin^2 B)(2\sin^2 C) \\=\bigl[3-(\cos 2A + \cos 2B + \cos 2C)\bigr]^2 \\{}\quad- 2\bigl[3-2(\cos 2A + \cos 2B + \cos 2C)+(\cos^2 2A + \cos^2 2B + \cos^2 2C)\bigr] \\{}\quad - 2(1-\cos 2A)(1-\cos 2B)(1-\cos 2C)$

$=9-6(\cos 2A + \cos 2B + \cos 2C) + (\cos 2A + \cos 2B + \cos 2C)^2 \\{}\quad - 6 + 4(\cos 2A + \cos 2B + \cos 2C) -2(\cos^2 2A + \cos^2 2B + \cos^2 2C) \\{}\quad - 2 + 2(\cos 2A + \cos 2B + \cos 2C) \\{}\quad- 2(\cos 2A\cos 2B + \cos 2B\cos 2C + \cos 2C\cos 2A) + 2\cos 2A\cos 2B\cos 2C\\=1 + (\cos 2A+\cos 2B+\cos 2C)^2 - 2(\cos2A\cos2B+\cos2B\cos2C+\cos2C\cos2A) \\{}\quad - 2(\cos^22A+\cos^22B+\cos^22C) + 2\cos2A\cos2B\cos2C$

$=1 + \cos^22A+\cos^22B+\cos^22C - 2(\cos^22A+\cos^22B+\cos^22C) + 2\cos2A\cos2B\cos2C \\=1 - (\cos^22A+\cos^22B+\cos^22C) + 2\cos2A\cos2B\cos2C \\=1 - \cos^22A-\cos^22B-\cos^2(2A+2B) + 2\cos2A\cos2B\cos2C$

$=1 - \cos^22A-\cos^22B-\bigl[\cos2A\cos2B-\sin2A\sin2B\bigr]^2 + 2\cos2A\cos2B\cos2C \\=1 - \cos^22A-\cos^22B + 2\cos2A\cos2B\cos2C \\{}\quad - \cos^22A\cos^22B - \sin^22A\sin^22B + 2\cos2A\cos2B\sin2A\sin2B \\=1 - \cos^22A-\cos^22B + 2\cos2A\cos2B(\cos2C+\sin2A\sin2B) \\{}\quad - \cos^22A\cos^22B - (1-\cos^22A)(1-\cos^22B) \\=1 - \cos^22A-\cos^22B -\cos^22A\cos^22B + 2\cos2A\cos2B(\cos2C+\sin2A\sin2B) \\{}\quad - (1-\cos^22A)(1-\cos^22B)$

$=1 - \cos^22A-\cos^22B -\cos^22A\cos^22B + 2\cos2A\cos2B(\cos2C+\sin2A\sin2B) \\{}\quad - 1 + \cos^22A + \cos^22B - \cos^22A\cos^22B \\=-2\cos^22A\cos^22B + 2\cos2A\cos2B(\cos2C+\sin2A\sin2B) \\=2\cos2A\cos2B(\cos2C+\sin2A\sin2B-\cos2A\cos2B) \\=2\cos2A\cos2B(\cos2C+\sin2A\sin2B-\cos2A\cos2B) \\=0,\ \text{since}\ \cos2C=\cos(2A+2B)=\cos2A\cos2B-\sin2A\sin2B.$