Inverse Trigonometric Functions Question 129
Question: The value $ 2{{\tan }^{-1}}[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} ] $ is equal to
Options:
A) $ {{\cos }^{-1}}( \frac{a\cos \theta +b}{a+b\cos \theta } ) $
B) $ {{\cos }^{-1}}( \frac{a+b\cos \theta }{a\cos \theta +b} ) $
C) $ {{\cos }^{-1}}( \frac{a\cos \theta }{a+b\cos \theta } ) $
D) $ {{\cos }^{-1}}( \frac{b\cos \theta }{a\cos \theta +b} ) $
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Answer:
Correct Answer: A
$ 2{{\tan }^{-1}}[ \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} ] $
$ ={{\cos }^{-1}}[ \frac{1-( \frac{a-b}{a+b} ){{\tan }^{2}}\frac{\theta }{2}}{1+( \frac{a-b}{a+b} ){{\tan }^{2}}\frac{\theta }{2}} ] $
$ [ \therefore 2{{\tan }^{-1}}x={{\cos }^{-1}}\frac{1-x^{2}}{1+x^{2}} ] $
$ ={{\cos }^{-1}}[ \frac{(a+b)-(a-b){{\tan }^{2}}\frac{\theta }{2}}{(a+b)+(a-b){{\tan }^{2}}\frac{\theta }{2}} ] $
$ ={{\cos }^{-1}}[ \frac{a( 1-{{\tan }^{2}}\frac{\theta }{2} )+b( 1+{{\tan }^{2}}\frac{\theta }{2} )}{a( 1+{{\tan }^{2}}\frac{\theta }{2} )+b( 1-{{\tan }^{2}}\frac{\theta }{2} )} ] $
$ ={{\cos }^{-1}}[ \frac{\frac{a( 1-{{\tan }^{2}}\frac{\theta }{2} )}{1+{{\tan }^{2}}\frac{\theta }{2}}+b}{a+b( \frac{1-{{\tan }^{2}}\frac{\theta }{2}}{1+{{\tan }^{2}}\frac{\theta }{2}} )} ] $
$ ={{\cos }^{-1}}[ \frac{a\cos \theta +b}{a+b\cos \theta } ] $
BETA
