Question

In a $△ABC\frac{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}{{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C}=2$
The triangle is

• A. equilateral
• B. obtuse
• C. isosceles
• D. right

Answer 1

The correct option is C isosceles

$\begin{array}{c}{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\left({\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C\right)\\ \Rightarrow 2+2\cos A\cdot \cos B\cdot \cos C=-4\cos A\cdot \cos B\cdot \cos C\\ \Rightarrow 6\cdot \cos A\cdot \cos B\cdot \cos C=0\\ \Rightarrow 3\left(2\cos A\cdot \cos B\right)\cdot \cos C=0\\ 3\cdot \cos \left(A+B\right)\cdot \cos \left(A+B\right)\cdot \cos C=0\\ \Rightarrow 3\cdot \cos \left(\pi -C\right)\cdot \cos C\cdot \cos \left(A-B\right)\\ \Rightarrow -3{\cos }^{2}C\cdot \cos \left(A-B\right)=0\\ C=\pi /2,\left(A-B\right)=\pi /2,A+B+C=\pi \\ hence,traingleisisosceles.\end{array}$