Appropriate Question\begin{gathered}\sf \: If \: tanA + sinA = m \: and \: tanA-sinA=n \\ \sf \: show \: that \: {m}^{2} \: - \: {n}^{2} \: = \: 4 \: \SQRT{mn} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} IftanA+sinA=mandtanA−sinA=nshowthatm 2 −n 2 =4 mn \large\underline{\sf{Solution-}} Solution− Given that,\rm :\longmapsto\:m = tanA + sinA:⟼m=tanA+sinAand\rm :\longmapsto\:n = tanA - sinA:⟼n=tanA−sinANow, Consider\rm :\longmapsto\: {m}^{2} - {n}^{2}:⟼m 2 −n 2 On substituting the VALUES of m and n, we get\rm \: = \: \: {(tanA + sinA)}^{2} - {(tanA - sinA)}^{2} = (tanA+sinA) 2 −(tanA−sinA) 2 We know\boxed{ \bf{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}} (x+y) 2 −(x−y) 2 =4xy So, using this identity, we get\rm \: = \: \: 4 \: tanA \: sinA = 4tanAsinA\bf\implies \: {m}^{2} - {n}^{2} = 4 \: tanA \: sinA - - - (1)⟹m 2 −n 2 =4tanAsinA−−−(1)Now, Consider\rm :\longmapsto\:4 \sqrt{mn}:⟼4 mn On substituting the values of m and n, we get\rm \: = \: \: 4 \sqrt{(tanA + sinA)(tanA - sinA)} = 4 (tanA+sinA)(tanA−sinA) We know,\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}} (x+y)(x−y)=x 2 −y 2 So, using this identity we get,\rm \: = \: \: 4 \sqrt{ {tan}^{2}A - {sin}^{2}A } = 4 tan 2 A−sin 2 A \rm \: = \: \: 4 \sqrt{ \dfrac{ {sin}^{2} A}{ {cos}^{2} A} - {sin}^{2}A } = 4 cos 2 Asin 2 A −sin 2 A \rm \: = \: \: 4 \sqrt{ \bigg(\dfrac{1}{ {cos}^{2} A} - 1\bigg){sin}^{2}A } = 4 ( cos 2 A1 −1)sin 2 A We know,\boxed{ \bf{ \: \frac{1}{cosx} = sinx}} cosx1 =sinx So, using this\rm \: = \: \: 4 \sqrt{ \bigg( {sec}^{2}A - 1\bigg){sin}^{2}A } = 4 (sec 2 A−1)sin 2 A We know,\boxed{ \bf{ \: {sec}^{2}x - {tan}^{2}x = 1}} sec 2 x−tan 2 x=1 So, using this, we get\rm \: = \: \: 4 \sqrt{ {tan}^{2} A \times {sin}^{2} A} = 4 tan 2 A×sin 2 A \rm \: = \: \: 4tanA \: sinA = 4tanAsinA\bf\implies \:4 \sqrt{mn} = 4 \: tanA \: sinA - - - (2)⟹4 mn =4tanAsinA−−−(2)From EQUATION (1) and (2), we concluded that\boxed{ \bf{ \: {m}^{2} - {n}^{2} = 4 \sqrt{mn}}} m 2 −n 2 =4 mn Additional Information:-Relationship between sides and T ratiossin θ = Opposite Side/Hypotenusecos θ = Adjacent Side/Hypotenusetan θ = Opposite Side/Adjacent Sidesec θ = Hypotenuse/Adjacent Sidecosec θ = Hypotenuse/Opposite Sidecot θ = Adjacent Side/Opposite SideReciprocal Identitiescosec θ = 1/sin θsec θ = 1/cos θcot θ = 1/tan θsin θ = 1/cosec θcos θ = 1/sec θtan θ = 1/COT θCo-function Identitiessin (90°−x) = cos xcos (90°−x) = sin xtan (90°−x) = cot xcot (90°−x) = tan xsec (90°−x) = cosec xcosec (90°−x) = sec xFundamental TRIGONOMETRIC Identitiessin²θ + cos²θ = 1sec²θ - tan²θ = 1cosec²θ - cot²θ = 1Explanation:#SAGARTHELEGEND