Mathematics - Properties of Triangle Question with Solution | TestHub

Explanation:

Given a triangle $△\mathrm{ABC}$, we have $\angle \mathrm{C}=120^{\circ }$ and $\cos \mathrm{A}+\cos \mathrm{B}=1$.

We know that the sum of angles in a triangle is $180^{\circ }$, so $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ }$.

Substituting $\mathrm{C}=120^{\circ }$, we get $\mathrm{A}+\mathrm{B}+120^{\circ }=180^{\circ }$, which implies $\mathrm{A}+\mathrm{B}=60^{\circ }$.

Now, let's use the given condition $\cos \mathrm{A}+\cos \mathrm{B}=1$.

We can use the sum-to-product formula for cosines: $\cos \mathrm{X}+\cos \mathrm{Y}=2\cos \left(\frac{\mathrm{X}+\mathrm{Y}}{2}\right)\cos \left(\frac{\mathrm{X}-\mathrm{Y}}{2}\right)$.

So, $2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1$.

Since $\mathrm{A}+\mathrm{B}=60^{\circ }$, we have $\frac{\mathrm{A}+\mathrm{B}}{2}=30^{\circ }$.

Substituting this value:

$2\cos (30^{\circ })\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1$

$2\left(\frac{\sqrt{3}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1$

$\sqrt{3}\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1$

$\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{\sqrt{3}}$

Now let's evaluate the options.

Option A: $\cos (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}}$

We know that $\cos (2\theta )=2{\cos }^2\theta -1$.

So, $\cos (\mathrm{A}-\mathrm{B})=2{\cos }^2\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-1$.

Substitute the value of $\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{\sqrt{3}}$:

$\cos (\mathrm{A}-\mathrm{B})=2{\left(\frac{1}{\sqrt{3}}\right)}^2-1=2\left(\frac{1}{3}\right)-1=\frac{2}{3}-1=-\frac{1}{3}$.

So, option A is incorrect.

Option C: $\cos (\mathrm{A}-\mathrm{B})=-\frac{1}{3}$

From our calculation above, $\cos (\mathrm{A}-\mathrm{B})=-\frac{1}{3}$.

So, option C is correct.

Option B: $|\cos \mathrm{A}-\cos \mathrm{B}|=\frac{\sqrt{6}}{3}$

We use the difference-to-product formula for cosines: $\cos \mathrm{X}-\cos \mathrm{Y}=-2\sin \left(\frac{\mathrm{X}+\mathrm{Y}}{2}\right)\sin \left(\frac{\mathrm{X}-\mathrm{Y}}{2}\right)$.

So, $\cos \mathrm{A}-\cos \mathrm{B}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$.

We know $\frac{\mathrm{A}+\mathrm{B}}{2}=30^{\circ }$, so $\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\sin (30^{\circ })=\frac{1}{2}$.

We have $\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{\sqrt{3}}$.

We can find $\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$ using the identity ${\sin }^2\theta +{\cos }^2\theta =1$.

${\sin }^2\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1-{\cos }^2\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1-{\left(\frac{1}{\sqrt{3}}\right)}^2=1-\frac{1}{3}=\frac{2}{3}$.

So, $\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\pm \sqrt{\frac{2}{3}}=\pm \frac{\sqrt{2}}{\sqrt{3}}=\pm \frac{\sqrt{6}}{3}$.

Therefore, $\cos \mathrm{A}-\cos \mathrm{B}=-2\left(\frac{1}{2}\right)\left(\pm \frac{\sqrt{6}}{3}\right)=\mp \frac{\sqrt{6}}{3}$.

Taking the absolute value: $|\cos \mathrm{A}-\cos \mathrm{B}|=\left|\mp \frac{\sqrt{6}}{3}\right|=\frac{\sqrt{6}}{3}$.

So, option B is correct.

Option D: ${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}-{\cos }^2\mathrm{C}=\frac{1}{4}$

We know $\mathrm{C}=120^{\circ }$, so $\cos \mathrm{C}=\cos (120^{\circ })=-\frac{1}{2}$.

${\cos }^2\mathrm{C}={\left(-\frac{1}{2}\right)}^2=\frac{1}{4}$.

We have $\cos \mathrm{A}+\cos \mathrm{B}=1$.

Square both sides: $(\cos \mathrm{A}+\cos \mathrm{B}{)}^2=1^2$

${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}+2\cos \mathrm{A}\cos \mathrm{B}=1$.

We need to find $\cos \mathrm{A}\cos \mathrm{B}$.

We know $\cos (\mathrm{A}+\mathrm{B})=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\sin \mathrm{B}$.

Since $\mathrm{A}+\mathrm{B}=60^{\circ }$, $\cos (\mathrm{A}+\mathrm{B})=\cos (60^{\circ })=\frac{1}{2}$.

So, $\frac{1}{2}=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\sin \mathrm{B}$.

This approach seems complicated. Let's try another way.

We have $\cos \mathrm{A}+\cos \mathrm{B}=1$.

We also know $\cos (\mathrm{A}-\mathrm{B})=-\frac{1}{3}$.

And $\cos (\mathrm{A}+\mathrm{B})=\cos (60^{\circ })=\frac{1}{2}$.

We know that $2\cos \mathrm{A}\cos \mathrm{B}=\cos (\mathrm{A}+\mathrm{B})+\cos (\mathrm{A}-\mathrm{B})$.

$2\cos \mathrm{A}\cos \mathrm{B}=\frac{1}{2}+\left(-\frac{1}{3}\right)=\frac{3-2}{6}=\frac{1}{6}$.

So, $\cos \mathrm{A}\cos \mathrm{B}=\frac{1}{12}$.

Now substitute this into ${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}+2\cos \mathrm{A}\cos \mathrm{B}=1$:

${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}+2\left(\frac{1}{12}\right)=1$

${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}+\frac{1}{6}=1$

${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}=1-\frac{1}{6}=\frac{5}{6}$.

Now, let's check the expression in option D:

${\cos }^2\mathrm{A}+{\cos }^2\mathrm{B}-{\cos }^2\mathrm{C}=\frac{5}{6}-\frac{1}{4}=\frac{10-3}{12}=\frac{7}{12}$.

Since $\frac{7}{12}\ne \frac{1}{4}$, option D is incorrect.

Therefore, the correct options are B and C.

The final answer is $\boxed{B,C}$.