Trigonometric Equations Question 102
Question: If a,b are different values of $ x $ satisfying $ a\cos x+b\sin x=c, $ then $ \tan ( \frac{\alpha +\beta }{2} )= $
[EAMCET 1986; Orissa JEE 2003]
Options:
A) $ a+b $
B) $ a-b $
C) $ \frac{b}{a} $
D) $ \frac{a}{b} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ a\cos x+b\sin x=c $
$ \Rightarrow $ $ a,( \frac{1-{{\tan }^{2}},(x/2)}{1+{{\tan }^{2}}(x/2)} )+\frac{2b\tan (x/2)}{1+{{\tan }^{2}}(x/2)}=c $
$ \Rightarrow $ $ (a+c){{\tan }^{2}}\frac{x}{2}-2b\tan \frac{x}{2}+(c-a)=0 $ This equation has roots $ \tan \frac{\alpha }{2} $ and $ \tan \frac{\beta }{2} $ .
$ \therefore $ $ \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{2b}{a+c} $ and $ \tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{c-a}{a+c} $ Now $ \tan ( \frac{\alpha }{2}+\frac{\beta }{2} )=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\frac{\frac{2b}{a+c}}{1-\frac{c-a}{a+c}}=\frac{b}{a} $ .
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