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\sqrt{a-b/a+b}=tanAtanB then show that (a-bCos2A)(a-bCos2B)=a^{2}-b^{2}
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.[1] To prove: (a-bCos2A)(a-bCos2B)=a^2-b^2 from eq.[1] 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...!!
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.[1]
To prove: (a-bCos2A)(a-bCos2B)=a^2-b^2
from eq.[1]
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...!!
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