Question

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

Options

• (a) 0 

• (b) −1 

• (c) 1

• (d) None of these 

Answer 1

(b)  −1
π = 180° 

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

We know that, 

\[\cos\left( \pi - \theta \right) = - cos\theta \text{ and } \sin\left( \pi - \theta \right) = sin\theta\]  

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

Now, using the identities 

\[\cos\left( A + B \right) = \cos A\cos B - \sin A\sin B\]  and \[\sin\left( A + B \right) = \sin A\cos B + \cos A\sin B\]

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

\[\Rightarrow \sec A\left( \cos B\cos C - \sin B\sin C \right) = \frac{- \cos A\left( \cos^2 B + \sin^2 B \right)}{\cos A}\]
\[ \Rightarrow \sec A\left( \cos B\cos C - \sin B\sin C \right) = \frac{- \cos A}{\cos A} = - 1\]