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# \sqrt{a-b/a+b}=tanAtanB then show that (a-bCos2A)(a-bCos2B)=a^{2}-b^{2}

9 years ago

sqrt{a-b/a+b}=tanAtanB  (given)

multiplying and dividing (a-b) in LHS...

sqrt{(a-b)(a-b)/(a+b)(a-b)}=tanAtanB

sqrt{(a-b)^2/a^2-b^2}=tanAtanB

squaring both sides...

(a-b)^2/a^2-b^2=tan^2Atan^2B

a^2-b^2=(a-b)^2/tan^2Atan^2B   ~eq.

To prove: (a-bCos2A)(a-bCos2B)=a^2-b^2

from eq.

we can prove,  (a-b)^2/tan^2Atan^2B=(a-bCos2A)(a-bCos2B)

RHS-

=> {a-b[(1-tan^2A)/(1+tan^2A)]}{a-b[(1-tan^2B)/(1+tan^2B)]}

=> [a-(b-btan^2A)/(1+tan^2A)][a-(b-btan^2B)/(1+tan^2B)]

=> [(a+atan^2A-b+btan^2A)(a+atan^2B-b+btan^2B)]/[1+tan^2A+tan^2B+tan^2Atan^2B]

on solving further....

=> (a^2+b^2-2ab)(tan^2B)/(tan^2Atan^4B)

=> (a-b)^2/tan^2Atan^2B

hence proved...!!