If tan|+sin|=m and tan|- sin|=n, show that (m2-n2)=4?mn

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Krishna Yeswanth

Tanx=(m+n)/2; sinx = (m-n)/2; Since cosec^2(x)-cot^2(x) = 1 (2/m-n)^2-(2/m+n)^2=1 (m^2-n^2)^2=16mn
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Writing angle l =θ, then question is -- ( tanθ + sinθ) = m and ( tanθ - sinθ) = n, show (m2 - n2) = 4√mn . now ( m+ n) = ( tanθ +sinθ +tanθ - sinθ ) = 2tanθ . now , ( m - n) =( tanθ + sinθ - tanθ +sinθ...
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Writing angle l =θ, then question is -- ( tanθ + sinθ) = m and ( tanθ - sinθ) = n, show (m2 - n2) = 4√mn . now ( m+ n) = ( tanθ +sinθ +tanθ - sinθ ) = 2tanθ . now , ( m - n) =( tanθ + sinθ - tanθ +sinθ ) = 2sinθ . so (m+n)×(m- n) = × 2sinθ OR ( m2 read less
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Writing angle l=β question is – (tanβ +sinβ) =m ,and (tanβ – sinβ) =n , to show (m2- n2) =4√ mn. Now , (m +n )= (tanβ +sinβ +tanβ – sinβ) = 2tanβ .now (m -n) = (tanβ +sinβ – tanβ+sinβ ) = 2 sinβ...
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Writing angle l=β question is – (tanβ +sinβ) =m ,and (tanβ – sinβ) =n , to show (m2- n2) =4√ mn. Now , (m +n )= (tanβ +sinβ +tanβ – sinβ) = 2tanβ .now (m -n) = (tanβ +sinβ – tanβ+sinβ ) = 2 sinβ . Now (m+n)(m-n) = (m2 – n2 ) =2tanβ.2sinβ = 4sinβtanβ . now , (mn) = (tanβ +sinβ)(tanβ- sinβ) = tan2β – sin2β = ( sin2β /cos2β ) - sin2β = ( sin2β -sin2β.cos2β)/cos2β ={ sin2β(1- cos2β)}/cos2β = Sin2β(sin2β/cos2β) = sin2β.tan2β = mn or sinβtanβ =√ mn . so (m2 - n2) = 4sinβtanβ = 4√ mn (proved). read less
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