Question

In triangle $ABC$, if $\frac{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}{{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C}=2$

then the triangle is

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

Answer 1

The correct option is A right-angled

$\frac{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}{{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C}=2$

$⟹{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2{\cos }^{2}A+2{\cos }^{2}B+2{\cos }^{2}C$

$⟹{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2(1-{\sin }^{2}A)+2(1-{\sin }^{2}B)+2(1-{\sin }^{2}C)$

$⟹3({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)=6$

$⟹{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

$⟹\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}=2$

$⟹3-(\cos 2A+\cos 2B+\cos 2C)=4$

$⟹\cos 2A+\cos 2B+\cos 2C=-1$

$A+B+C=\pi ⟹2C=2\pi -2(A+B)$

Hence, $2cos\left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)+\cos (2\pi -2(A+B\left)\right)=-1$

$⟹2\cos (A+B)2\cos (A-B)+\cos 2(A+B)=-1$

$⟹2\cos (A+B)\cos (A-B)+2{\cos }^2(A+B)-1=-1$

$⟹2\cos (A+B)\left[\cos \right(A-B)+\cos (A+B\left)\right]=0$

$⟹2\cos (\pi -C)\times 2\cos A\cos (-B)=0$

$⟹4\cos A\cos B\cos C=0$

$⟹A={90}^o$ or $B={90}^o$ or $C={90}^o$

Hence, $\Delta ABC$ is a right triangle.