Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$

Question

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!

Answer 1

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.

Answer 2

From where you have left off,

$$\sin^2A+\sin^2B+\sin^2C=2\implies1-\cos^2A+1-(\cos^2B-\sin^2C)=2$$

$$\implies\cos^2A+\cos^2B-\sin^2C=0$$

Using $\displaystyle\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$

$$\implies\cos^2A+\cos^2B-\sin^2C=\cos^2A+\cos(B+C)\cos(B-C)$$

As $\displaystyle A+B+C=\pi, B+C=\pi-A,\cos(B+C)=\cos(\pi-A)=-\cos A,$

$$\implies\cos^2A+\cos(B+C)\cos(B-C)=\cos^2A-\cos A\cos(B-C)$$

$$=\cos A\left[\cos A -\cos(B-C)\right]$$

$$=\cos A\left[-\cos(B+C) -\cos(B-C)\right]$$

$$=\cos A\left[-2\cos B\cos C\right]$$