Question

The value of ${\cos }^2\left(A-B\right)+{\cos }^2\left(B\right)-2\cos \left(A-B\right)\cos \left(A\right)\cos \left(B\right)$ is

• A. $\sin \left(A\right)$
• B. ${\sin }^2\left(A\right)$
• C. ${\cos }^2\left(A\right)$
• D. $\cos \left(A\right)$

Answer 1

The correct option is B

${\sin }^2\left(A\right)$

Explanation for correct option

The value of ${\cos }^2\left(A-B\right)+{\cos }^2\left(B\right)-2\cos \left(A-B\right)\cos \left(A\right)\cos \left(B\right)$

$$\begin{gathered} =\cos \left(A-B\right)\left[\cos \left(A-B\right)-2\cos \left(A\right)\cos \left(B\right)\right]+{\cos }^2\left(B\right) \\ =\cos \left(A-B\right)\left[\cos \left(A\right)\cos \left(B\right)+\sin \left(A\right)\sin \left(B\right)-2\cos \left(A\right)\cos \left(B\right)\right]+{\cos }^2\left(B\right)\left[\because \cos \left(A-B\right)=\cos \left(A\cos \left(B\right)\right)+\sin \left(A\right)\sin \left(B\right)\right] \\ =\cos \left(A-B\right)\left[\sin \left(A\right)\sin \left(B\right)-\cos \left(A\right)\cos \left(B\right)\right]+{\cos }^2\left(B\right) \\ =\cos \left(A-B\right)\left[-\cos \left(A+B\right)\right]+{\cos }^2\left(B\right) \\ =-\frac{1}{2}\left[2\cos \left(A+B\right)\cos \left(A-B\right)\right]+{\cos }^2\left(B\right) \\ =-\frac{1}{2}\left[\cos \left(2A\right)+\cos \left(2B\right)\right]+{\cos }^2\left(B\right) \\ =-\frac{1}{2}\left[2{\cos }^2\left(A\right)-1+2{\cos }^2\left(B\right)-1\right]+{\cos }^2\left(B\right) \\ =1-{\cos }^2\left(B\right)-{\cos }^2\left(A\right)+{\cos }^2\left(B\right) \\ =1-{\cos }^2\left(A\right) \\ ={\sin }^2\left(A\right) \end{gathered}$$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$