If `y=2/(sqrt(a^2-b^2)){tan^(-1)((a-b)/(a+b))tanx/2}` , then show that `(d^2y)/(dx^2)=(bsinx)/((a+bcosx)^2)`

Explanation

`Rightarrow quad frac{d y}{d x}=frac{2}{sqrt{a^{2}-b^{2}}} frac{1}{1+(frac{a-b}{a+b}) tan ^{2}(frac{x}{2})}`
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cdot sqrt{frac{a-b}{a+b}} cdot frac{1}{2} sec ^{2}(frac{x}{2})`
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Rightarrow quad frac{d y}{d x}=frac{2}{sqrt{a^{2}-b^{2}}} cdot frac{(a+b)}{(a+b)+(a-b) tan ^{2}(frac{x}{2})}`
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cdot sqrt{frac{a-b}{a+b}} cdot frac{1}{2} sec ^{2}(frac{x}{2})`
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Rightarrow quad frac{d y}{d x}=sqrt{frac{1}{(a^{2}-b^{2})} times frac{a-b}{a+b} times(a+b)^{2}}`
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frac{(1+tan ^{2}(frac{x}{2}))}{.{n}^{2}(frac{x}{2}))+b(1-tan ^{2}(frac{x}{2}))}`
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Rightarrow frac{d y}{d x}=frac{1}{a+b(frac{1-tan ^{2}(frac{x}{2})}{1+tan ^{2}(frac{x}{2})})}`
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Rightarrow quad frac{d y}{d x}=frac{1}{a+b cos x}`
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