(c) यहाँ  $\sin (\alpha – \beta ) = \sin (\theta – \beta – \overline {\theta – \alpha } )$

$ = \sin (\theta – \beta )\cos (\theta – \alpha ) – \cos (\theta – \beta )\sin (\theta – \alpha )$

$ = ba – \sqrt {1 – {b^2}} \sqrt {1 – {a^2}} $

एवं  $\cos (\alpha – \beta ) = \cos (\theta – \beta – \overline {\theta – \alpha } )$

$ = \cos (\theta – \beta )\cos (\theta – \alpha ) + \sin (\theta – \beta )\sin (\theta – \alpha )$

$ = a\sqrt {1 – {b^2}} + b\sqrt {1 – {a^2}} $

$\therefore $ अतः दिया गया व्यंजक ${\cos ^2}(\alpha – \beta ) + 2ab\sin (\alpha – \beta )$

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

$ = {a^2} + {b^2}$.

ट्रिक :  $\alpha = 30^\circ ,\beta = 60^\circ ,$$\theta = 90^\circ $ रखने पर

$a = \frac{1}{2},b = \frac{1}{2}$

$\therefore {\cos ^2}(\alpha – \beta ) + 2ab\sin (\alpha – \beta ) = \frac{3}{4} + \frac{1}{2} \times \left( { – \frac{1}{2}} \right) = \frac{1}{2}$

जो कि विकल्प $(c)$ द्वारा दिया जाता है।