A) \[{{\cos }^{-1}}\left( \frac{a\cos \theta +b}{a+b\cos \theta } \right)\]
B) \[{{\cos }^{-1}}\left( \frac{a+b\cos \theta }{a\cos \theta +b} \right)\]
C) \[{{\cos }^{-1}}\left( \frac{a\cos \theta }{a+b\cos \theta } \right)\]
D) \[{{\cos }^{-1}}\left( \frac{a\cos +b\theta }{a+b\cos \theta } \right)\]
Correct Answer: A
Solution :
\[2{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} \right]\]\[={{\cos }^{-1}}\left[ \frac{1-\left( \frac{a-b}{a+b} \right){{\tan }^{2}}\frac{\theta }{2}}{1+\left( \frac{a-b}{a+b} \right){{\tan }^{2}}\frac{\theta }{2}} \right]\] \[\left( \because 2{{\tan }^{-1}}x={{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] \[={{\cos }^{-1}}\left[ \frac{(a+b)-(a-b){{\tan }^{2}}\frac{\theta }{2}}{(a+b)+(a-b){{\tan }^{2}}\frac{\theta }{2}} \right]\] \[={{\cos }^{-1}}\left[ \frac{a\left( 1-{{\tan }^{2}}\frac{\theta }{2} \right)+b\left( 1+{{\tan }^{2}}\frac{\theta }{2} \right)}{a\left( 1+{{\tan }^{2}}\frac{\theta }{2} \right)+b\left( 1-{{\tan }^{2}}\frac{\theta }{2} \right)} \right]\] \[={{\cos }^{-1}}\left[ \frac{\frac{a\left( 1-{{\tan }^{2}}\frac{\theta }{2} \right)}{1+{{\tan }^{2}}\frac{\theta }{2}}+b}{a+b\left( \frac{1-{{\tan }^{2}}\frac{\theta }{2}}{1+{{\tan }^{2}}\frac{\theta }{2}} \right)} \right]={{\cos }^{-1}}\left[ \frac{a\cos \theta +b}{a+b\cos \theta } \right]\].
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