Question

If $A+B=C$, then $co{s}^{2}A+co{s}^{2}B+co{s}^{2}C-2cosAcosBcosC$ is equal to

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Find the value of $co{s}^{2}A+co{s}^{2}B+co{s}^{2}C-2cosAcosBcosC$:

Given, $A+B=C$

Take “cos” on both sides, we get

$\cos (A+B)=\cos C$

Similarly, $\sin (A+B)=\sin C$

Now,

${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C–2\cos A\cos B\cos C$

$$\begin{gathered} =\frac{(1+\cos 2A)}{2}+\frac{(1+\cos 2B)}{2}+\frac{(1+\cos 2C)}{2}–2\cos A\cos B\cos C \\ =\frac{3}{2}+\left(\frac{1}{2}\right)(\cos 2A+\cos 2B+\cos 2C)–2\cos A\cos B\cos C \\ =\frac{3}{2}+\left(\frac{1}{2}\right)\left[2\cos (A+B)\cos (A–B)+\cos 2C\right]–\left[\cos (A+B)+\cos (A–B)\right]\cos C \\ =\frac{3}{2}+\cos (A+B)\cos (A–B)+\frac{\cos 2C}{2}–{\cos }^{2}C–\cos (A–B)\cos C \\ =\frac{3}{2}+\cos C\cos (A-B)+\frac{\cos 2C}{2}–{\cos }^{2}C–\cos (A–B)\cos C \\ =\frac{3}{2}+\frac{\left(\cos 2C–2{\cos }^{2}C\right)}{2} \\ =\frac{3}{2}+\frac{(2{\cos }^{2}C–1–2{\cos }^{2}C)}{2} \\ =\frac{3}{2}–\frac{1}{2} \\ =\frac{(3–1)}{2} \\ =\frac{2}{2} \\ =\mathbf{1} \end{gathered}$$

Hence, Option ‘A’ is Correct.