Question
If in a $\triangle\text{ABC},\cos^2\text{A}+\cos^2\text{B}+\cos^2\text{C}=1,$ prove that the triangle is right angled.
✓
Answer
Let ABC be any triangle.
In $\triangle\text{ABC},$
$\cos^2\text{A}+\cos^2\text{B}+\cos^2\text{C}=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2[\pi-(\text{B + A})]=1$ $(\because\text{A + B + C} = \pi)$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2(\text{B + A})=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=1-\cos^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B})^2$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2\text{A}\cos^2\text{B}+\cos^2\text{A}\sin^2\text{B}\\+2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos^2\text{A}(1-\sin^2\text{B})+\cos^2\text{B}(1-\sin^2\text{A})$ $=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow2\cos^2\text{A}\cos^2\text{B}=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=0$
$\Rightarrow\cos(\text{A + B)}=0$
$\Rightarrow\cos\text{(A + B)}=\cos90^{\circ}$
$\Rightarrow\text{A + B}=90^{\circ}$
$\Rightarrow\text{C}=90^{\circ}$ $(\because\text{A + B + C }= 180^{\circ})$
Hence, $\triangle\text{ABC}$ is right angled.
In $\triangle\text{ABC},$
$\cos^2\text{A}+\cos^2\text{B}+\cos^2\text{C}=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2[\pi-(\text{B + A})]=1$ $(\because\text{A + B + C} = \pi)$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2(\text{B + A})=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=1-\cos^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B})^2$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2\text{A}\cos^2\text{B}+\cos^2\text{A}\sin^2\text{B}\\+2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos^2\text{A}(1-\sin^2\text{B})+\cos^2\text{B}(1-\sin^2\text{A})$ $=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow2\cos^2\text{A}\cos^2\text{B}=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=0$
$\Rightarrow\cos(\text{A + B)}=0$
$\Rightarrow\cos\text{(A + B)}=\cos90^{\circ}$
$\Rightarrow\text{A + B}=90^{\circ}$
$\Rightarrow\text{C}=90^{\circ}$ $(\because\text{A + B + C }= 180^{\circ})$
Hence, $\triangle\text{ABC}$ is right angled.
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