Trigonometric Equations Question 192
Question: In a $ \Delta ABC $ , if $ b^{2}+c^{2}=3a^{2} $ , then $ \cot B+\cot C-\cot A= $
[MP PET 1991]
Options:
A) 1
B) $ \frac{ab}{4\Delta } $
C) 0
D) $ \frac{ac}{4\Delta } $
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Answer:
Correct Answer: C
Solution:
- $ \cot B+\cot C-\cot A=\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}-\cot A $ $ =\frac{\sin C\cos B+\cos C\sin B}{\sin B\sin C}-\cot A $ $ =\frac{\sin (B+C)}{\sin B\sin C}-\frac{\cos A}{\sin A} $ $ =\frac{{{\sin }^{2}}A-\sin B\sin C\cos A}{\sin A\sin B\sin C}=\frac{a^{2}-bc\cos A}{k(abc)} $ Since $ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k $ (say) and $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} $ $ =\frac{a^{2}-bc\frac{(b^{2}+c^{2}-a^{2})}{2bc}}{(abc)k} $ $ =\frac{(a^{2}-a^{2})}{abc,k}=0,{ As\frac{b^{2}+c^{2}-a^{2}}{2}=\frac{3a^{2}-a^{2}}{2}=\frac{2a^{2}}{2}=a^{2} }. $
BETA
