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प्रश्न
Prove that `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1((a costheta+b)/(a+b costheta))`
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उत्तर
LHS = `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1{(1-(sqrt((a-b)/(a+b))tan theta/2)^2)/(1+(sqrt((a-b)/(a+b))tan theta/2)^2)}` `[because2tan^-1(x)=cos^-1{(1-x^2)/(1+x^2)}]`
`=cos^-1{(1-(a-b)/(a+b)tan^2 theta/2)/(1+(a-b)/(a+b)tan^2 theta/2)}`
`=cos^-1{((a+b)-(a-b)tan^2 theta/2)/((a+b)+(a-b)tan^2 theta/2)}`
`=cos^-1{(a+b-atan^2 theta/2+btan^2 theta/2)/(a+b+atan^2 theta/2-btan^2 theta/2)}`
`=cos^-1{(a(1-tan^2 theta/2)+b(1+tan^2 theta/2))/(a(1+tan^2 theta/2)+b(1-tan^2 theta/2))}`
`=cos^-1{(a((1-tan^2 theta/2)/(1+tan^2 theta/2))+b((1+tan^2 theta/2)/(1+tan^2theta/2)))/(a((1+tan^2 theta/2)/(1+tan^2 theta/2))+b((1-tan^2 theta/2)/(1-tan^2 theta/2)))}` `["Dividing" N^r and D^r by 1+tan^2 theta/2]`
`=cos^-1{(a((1-tan^2 theta/2)/(1+tan^2 theta/2))+b)/(a+b((1-tan^2 theta/2)/(1-tan^2 theta/2)))}`
`=cos^-1{(acos theta+b)/(a+bcostheta)}`=RHS
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