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

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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 

MCQ
Short/Brief Note
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Solution

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

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

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

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Chapter 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.4 [Page 27]

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R.D. Sharma Mathematics [English] Class 11
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.4 | Q 2 | Page 27

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