The value \[2{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} \right]\]is equal to

\[{{\cos }^{-1}}\left( \frac{a+b\cos \theta }{a\cos \theta +b} \right)\]

\[{{\cos }^{-1}}\left( \frac{a\cos \theta +b}{a+b\cos \theta } \right)\]

\[{{\cos }^{-1}}\left( \frac{a\cos \theta }{a+b\cos \theta } \right)\]

\[{{\cos }^{-1}}\left( \frac{b\cos \theta }{a\cos \theta +b} \right)\]

The value \[2{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\th

1.	The value \[2{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} \right]\]is equal to
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.	\[{{\cos }^{-1}}\left( \frac{b\cos \theta }{a\cos \theta +b} \right)\]
Answer» B. \[{{\cos }^{-1}}\left( \frac{a+b\cos \theta }{a\cos \theta +b} \right)\]

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The value \[2{{\tan }^{-1}}\left[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} \right]\]is equal to

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