Question

If $\cos (\alpha -\theta )=a$, $\cos (\beta -\theta )=b$, then the value of ${\sin }^2(\alpha -\beta )+2ab\cos (\alpha -\beta )$ is:

• A. $a^2+b^2$
• B. $a^2-b^2$
• C. $b^2-a^2$
• D. $-a^2-b^2$

Answer 1

The correct option is A $a^2+b^2$

Given that,

$\cos (\alpha -\theta )=a⟹\sin (\alpha -\theta )=\sqrt{1-a^2}$

$\cos (\beta -\theta )=b⟹\sin (\beta -\theta )=\sqrt{1-b^2}$

Now, ${\sin }^2(\alpha -\beta )+2ab\cos (\alpha -\beta )$

$=1-{\cos }^2(\alpha -\beta )+2ab\cos (\alpha -\beta )$

$=1-{\cos }^2\left[\right(\alpha -\theta )-(\beta -\theta \left)\right]+2ab\cos \left[\right(\alpha -\theta )-(\beta -\theta \left)\right]$

$=1-{\left[\cos \right(\alpha -\theta \left)\cos \right(\beta -\theta )+\sin (\alpha -\theta \left)\sin \right(\beta -\theta \left)\right]}^2+2ab\left[\cos \right(\alpha -\theta \left)\cos \right(\beta -\theta )+\sin (\alpha -\theta \left)\sin \right(\beta -\theta \left)\right]$

$=1-{[ab+\sqrt{1-a^2}\sqrt{1-b^2}]}^2+2ab[ab+\sqrt{1-a^2}\sqrt{1-b^2}]$

$=1-a^2{b}^2-(1-a^2)(1-b^2)-2ab\sqrt{1-a^2}\sqrt{1-b^2}+2a^2{b}^2+2ab\sqrt{1-a^2}\sqrt{1-b^2}$

$=1-(1-a^2)(1-b^2)+a^2{b}^2$

$=1-1+a^2+b^2-a^2{b}^2+a^2{b}^2$

$=a^2+b^2$ (Option A)