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Show that the triangle whose angles satisfy the equality $$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$ is right-angled.

I've tried many times, but was unsuccessful.

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    $\begingroup$ Welcome to MSE. Your questions will have more chances to be answered rather than downvoted if you show your work: what have you tried? And, instead of giving a link to a .jpg file, please use MathJax $\endgroup$ – user539887 May 10 '18 at 20:58
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Hint: Show that \begin{eqnarray*} 1-\cos^2 A-\cos^2 B-\cos^2C+ 2\cos A \cos B \cos C =0 \end{eqnarray*} for any triangle.

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\begin{align} \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} &=2 . \end{align}

There are known identities for any $\triangle ABC$ with sides $a,b,c$, angles $A,B,C$ radius of inscribed circle $r$, radius of circumscribed circle $R$ and semiperimeter $\rho=\tfrac12(a+b+c)$:

\begin{align} a^2+b^2+c^2 &= 2\rho^2-8r\,R-2r^2 ,\\ \cos A \cos B \cos C &=\frac{\rho^2-(r+2R)^2}{4R^2} . \end{align}

\begin{align} {\sin^2 A + \sin^2 B + \sin^2 C} &=2(\cos^2 A + \cos^2 B + \cos^2 C) ,\\ {\sin^2 A + \sin^2 B + \sin^2 C} &=2(1-\sin^2 A + 1-\sin^2 B + 1-\sin^2 C) ,\\ 3(\sin^2 A + \sin^2 B + \sin^2 C) &=6 ,\\ \sin^2 A + \sin^2 B + \sin^2 C &=2 ,\\ 4R^2\sin^2 A + 4R^2\sin^2 B + 4R^2\sin^2 C &=2\cdot4R^2 ,\\ a^2+b^2+c^2&=8R^2 ,\\ 2\rho^2-8r\,R-2r^2&=8R^2 ,\\ \rho^2-(r+2R)^2&=0 ,\\ \frac{\rho^2-(r+2R)^2}{4R^2} =0 . \end{align}

And since

\begin{align} \cos A \cos B \cos C &=\frac{\rho^2-(r+2R)^2}{4R^2} , \end{align}

one of $\cos A,\cos B,\cos C$ must be 0, hence, one of the angles must be $\tfrac\pi2$.

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COMMENT.- First note that if $C=90^{\circ}$ then you have $\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\circ}$ you can consider that $\sin C=\sin (180^{\circ}-(A+B))=\sin(A+B)$ and $\cos C= \cos (180^{\circ}-(A+B))=-\cos(A+B)$ so you get $$\sin^2 A+\sin^2 B+\sin^2 (A+B)=2(\cos^2A+\cos ^2 B+\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish.

EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\sin^2 A+\sin^2 B+\sin^2 (A+B) = 2\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\dfrac{\pi}{2}$ (the "if" is obvious and we need the "only if"). So put $A + B = \dfrac{\pi}{2}\pm h$ with $h\ne 0$ and look at the functions $$\sin^2(x)+\sin^2(\dfrac{\pi}{2}\pm h-x)+\sin^2(\dfrac{\pi}{2}\pm h),\quad0\lt x\lt\dfrac{\pi}{2}$$ which becomes for $h$ positive and $h$ negative respectively $$f_1(x)=\sin^2(x)+\cos^2(x-h)+\cos^2(h)\qquad (1)\\f_2(x)=\sin^2(x)+\cos^2(x+h)+\cos^2(h)\qquad (2)$$ Now you can prove that $$\begin{equation}f_1(x)\begin{cases}\lt2\text{ when } 0\lt x\lt h\\=2 \text { when }x=h \space\text {discarded because } B\ne\dfrac{\pi}{2} \\\gt 2\text { when }h\lt x\lt\dfrac{\pi}{2}\\ \end{cases}\end{equation}$$

and that $f_2(x)$ is smaller than $2$ on $0\lt x\lt \dfrac{\pi}{2}$.

This requires elementary calculus that I guess you know how to handle.

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  • $\begingroup$ I can't get it please elaborate. $\endgroup$ – user390410 May 10 '18 at 21:37
  • $\begingroup$ Look at my edition later,please, where I reply your comment. Regards. $\endgroup$ – Piquito May 11 '18 at 11:54
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hint: try to get the identity ${\sin^2(x)} + {\cos^2(x)} =1$

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