<s>[INST] Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.  
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 C)$
$3(\sin^2 A+\sin^2 B+\sin^2 C)=6$
$(\sin^2 A+\sin^2 B+\sin^2 C)=2$
What to do now? Please help!
 [/INST] The last equation gives: cos(2A) + cos(2B) + cos(2C) = - 1 => 2cos(A+B)cos(A-B) + 2(cos(C))^2 = 0 ==> -2cos(C)cos(A-B) + 2(cos(C))^2 = 0 ==> 2cos(C)(-cos(A-B) + cos(C)) = 0. So cos(C) = 0 gives C = pi/2 or cos(A-B) = cos(C). So A-B = C or -C and this means A = B + C or B = A + C so 2A = pi so A =pi/2 or 2B = pi so B = pi/2. So either way A or B  or C = pi/2. ABC is a right triangle.
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