Trigonometric Equations Question 195
Question: In $ \Delta ABC,( \cot \frac{A}{2}+\cot \frac{B}{2} ),( a{{\sin }^{2}}\frac{B}{2}+b{{\sin }^{2}}\frac{A}{2} ) $ =
[Roorkee 1988]
Options:
A) $ \cot C $
B) $ c\cot C $
C) $ \cot \frac{C}{2} $
D) $ c\cot \frac{C}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ { \cot \frac{A}{2}+\cot \frac{B}{2} }{ a{{\sin }^{2}}\frac{B}{2}+b{{\sin }^{2}}\frac{A}{2} } $ = $ { \frac{\cos \frac{C}{2}}{\sin \frac{A}{2}\sin \frac{B}{2}} }\text{ }{ a{{\sin }^{2}}\frac{B}{2}+b{{\sin }^{2}}\frac{A}{2} } $ = $ { \cos \frac{C}{2} }{ a\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}+b\frac{\sin \frac{A}{2}}{\sin \frac{B}{2}} } $ $ =\sqrt{\frac{s(s-c)}{ab}}{ a\frac{\sqrt{\frac{(s-a)(s-c)}{ac}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}+b\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}} } $ $ =\sqrt{\frac{s(s-c)}{ab}}{ \sqrt{( \frac{s-a}{s-b} )ab}+\sqrt{( \frac{s-b}{s-a} )ab} } $ = $ \sqrt{s(s-c)}{ \frac{s-a+s-b}{\sqrt{(s-a)(s-b)}} }=\sqrt{s(s-c)}{ \frac{2s-a-b}{\sqrt{(s-a)(s-b)}} } $ $ =c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=c\cot \frac{C}{2} $ . Trick: Such type of unconditional problems can be checked by putting the particular values for $ a=1, $ $ b=\sqrt{3}, $ $ c=2 $ and $ A=30^{o} $ , $ B=60^{o},C=90^{o} $ . Hence expression is equal to 2 which is given by (d).
BETA
