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y=2/Sqrt(a^2-b^2)[tan^(-1){sqrt(a-b/a+b)tan(x/2)}] show that dy/dx=1/(a+bcosx)

y=2/Sqrt(a^2-b^2)[tan^(-1){sqrt(a-b/a+b)tan(x/2)}]
show that dy/dx=1/(a+bcosx)

Grade:12th pass

1 Answers

Sheeba
11 Points
6 years ago
dy/dx=(2/(sqrt(a2-b2)))*((1/(1+((a-b)/(a+b))tan2(x/2))))*(sqrt((a-b)/(a+b)))*sec2x/2*1/2
cancel 2 and sqrt(a-b)  we get
dy/dx=(1/(1+((a-b)/(a+b)tan2x/2))*(1/(a+b))*sec2x/2
taking sec2x/2 with denominator
dy/dx=(1/cos2x/2+(((a-b)/(a+b))*sin2x/2))
dy/dx=(1/acos2x/2+bcos2x/2+asin2x/2-bsin2x/2)
dy/dx=1/a(cos2x/2+sin2x/2)+b(cos2x/2-sin2x/2)
dy/dx=1/(a+bcosx)

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