Question

If A + B + C = 180°, then
sec A (cos B cos C − sin B sin C) is equal to

(a) 0
(b) −1
(c) 1
(d) None of these

Answer 1

(b) −1

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

We know that, $\cos \left(\mathrm{\pi }-\mathrm{\theta }\right)=-\mathrm{cos\theta }\mathrm{and}\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }$,

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

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

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

$$\begin{gathered} \Rightarrow \sec A\left(\cos B\cos C-\sin B\sin C\right)=\frac{-\cos A\left({\cos }^{2}B+{\sin }^{2}B\right)}{\cos A} \\ \Rightarrow \sec A\left(\cos B\cos C-\sin B\sin C\right)=\frac{-\cos A}{\cos A}=-1 \end{gathered}$$