Question

If $\cos (\theta -\alpha )=a$ and $\sin (\theta -\beta )=b\left(0\right)<\theta -\alpha ,\theta -\beta <\frac{\pi }{2})$, then ${\cos }^2(\alpha -\beta )+2ab\sin (\alpha -\beta )$ is equal to

• A. $4a^2{b}^2$
• B. $a^2-b^2$
• C. $a^2+b^2$
• D. $-a^2{b}^2$

Answer 1

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

We have
$\sin (\alpha -\beta )=\sin (\alpha -\theta +\theta -\beta )$
$=\sin \left[\right(\theta -\beta )-(\theta -\alpha \left)\right]$
$=\sin (\theta -\beta )\cos (\theta -\alpha )-\cos (\theta -\beta )\sin (\theta -\alpha )$
$=ba-\sqrt{1-b^2}\sqrt{1-a^2}$
and $\cos (\alpha -\beta )=\cos \left[\right(\theta -\beta )-(\theta -\alpha \left)\right]$
$=\cos (\theta -\beta )\cos (\theta -\alpha )+\sin (\theta -\beta )\sin (\theta -\alpha )$
$=a\sqrt{1-b^2}+b\sqrt{1-a^2}$
Substituting these values in the given expression, we get
${\cos }^2(\alpha -\beta )+2ab\sin (\alpha -\beta )$
$=(a\sqrt{1-b^2}+b\sqrt{1-a^2})+2ab[ab-\sqrt{(1-a^2)}\sqrt{(1-b^2)}]$
$=a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{(1-a^2)}\sqrt{(1-b^2)}+2a^2{b}^2-2ab\sqrt{(1-a^2)}\sqrt{(1-b^2)}=a^2+b^2$.