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If $ y=\frac{2}{\Large\sqrt {a^2-b^2}}\tan^{-1}\bigg[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} \bigg],$ prove that $ \Large\frac{dy}{dx}=\frac{1}{a+b \cos x}, \normalsize a>b>0 $

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Toolbox:
  • $ cosx=\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}}$
  • Use chain rule and differentiate
$ \large\frac{dy}{dx}=\large\frac{2}{\sqrt{a^2-b^2}}$x$ \large\frac{1}{1+\large\frac{a-b}{a+b}tan^2\large\frac{x}{2}}$x$ \sqrt{\large\frac{a-b}{a+b}}sec^2\large\frac{x}{2}$x$\large\frac{1}{2}$
 
Simplify which we get
$ \large\frac{dy}{dx}=\large\frac{1+tan^2\large\frac{x}{2}}{a(1+tan^2\large\frac{x}{2})+b(1-tan^2\large\frac{x}{2})}$
 
Devide number and then by $ 1+tan^2\large\frac{x}{2}$
$ \large\frac{dy}{dx}=\large\frac{1}{a+b\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}}}=\large\frac{1}{a+b\: cosx}$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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