If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to

(b) −1

$\pi=180^{\circ}$

$\sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (\pi-(A+B))-\sin B \sin (\pi-(A+B))}{\cos A}$

We know that, $\cos (\pi-\theta)=-\cos \theta$ and $\sin (\pi-\theta)=\sin \theta$,

$\therefore \sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (A+B)-\sin B \sin (A+B)}{\cos A}$

Now, using the identities $\cos (A+B)=\cos A \cos B-\sin A \sin B$ and $\sin (A+B)=\sin A \cos B+\cos A \sin B$, we get

$\sec A(\cos B \cos C-\sin B \sin C)=\frac{-\cos A \cos B^{2}+\cos B \sin A \sin B-\sin B \sin A \cos B-\sin ^{2} B \cos A}{\cos A}$

$\Rightarrow \sec A(\cos B \cos C-\sin B \sin C)=\frac{-\cos A\left(\cos ^{2} B+\sin ^{2} B\right)}{\cos A}$

$\Rightarrow \sec A(\cos B \cos C-\sin B \sin C)=\frac{-\cos A}{\cos A}=-1$