Trigonometric Equations Question 146
Question: In a triangle $ PQR $ , $ \angle R=\frac{\pi }{2}. $ If $ \tan ( \frac{P}{2} ) $ and $ \tan ( \frac{Q}{2} ) $ are the roots of the equation $ ax^{2}+bx+c=0(a\ne 0). $ then
[IIT 1999; MP PET 2000; AIEEE 2005]
Options:
A) $ a+b=c $
B) $ b+c=a $
C) $ a+c=b $
D) $ b=c $
Show Answer
Answer:
Correct Answer: A
Solution:
- Given $ \tan \frac{P}{2}+\tan \frac{Q}{2}=-\frac{b}{a} $ and $ \tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a} $ or $ \tan \frac{\alpha }{2}+\tan ( \frac{\pi }{4}-\frac{\alpha }{2} )=-\frac{b}{a} $ , $ \tan \frac{\alpha }{2}.\tan ( \frac{\pi }{4}-\frac{\alpha }{2} )=\frac{c}{a} $
$ \therefore $ $ \tan \frac{\alpha }{2}+\frac{1-\tan \frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a} $
$ \Rightarrow $ $ \frac{{{\tan }^{2}}\frac{\alpha }{2}+1}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a} $ ?..(i) Similarly, $ \frac{\tan \frac{\alpha }{2}( 1-\tan \frac{\alpha }{2} )}{1+\tan \frac{\alpha }{2}}=\frac{c}{a} $
$ \Rightarrow $ $ \frac{\tan \frac{\alpha }{2}-{{\tan }^{2}}\frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=\frac{c}{a} $ ……(ii) By adding (i) and (ii), we get $ \frac{1+\tan \frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=-\frac{b}{a}+\frac{c}{a} $
$ \Rightarrow $ $ -b+c=a $
$ \Rightarrow $ $ c=a+b $ .
BETA
