Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTRIGONOMETRIC FUNCTIONS1 Mark
MCQ
In any $\triangle\text{ABC},\text{a}(\text{b}\cos\text{C}-\text{c}\cos\text{B})=$
A
$a^2$
✓
$b^2-c^2$
C
$0$
D
$b^2+c^2$
✓
Answer
Correct option: B.
$b^2-c^2$
$b^2-c^2$
Solution:
Using cosine rule, we have
$\text{a(b}\cos\text{C}-\text{c}\cos\text{B})$
$=\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)$
$=\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{c}^2-\text{a}^2+\text{b}^2}{2}$
$=\frac{2\text{b}^2-2\text{c}^2}{2}$
$=\text{b}^2-\text{c}^2$
Hence, the correct answer is option (b).
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