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If tanθ + sinθ = m and tanθ – sinθ = n, show that m^2–n^2=4√(mn). - Mathematics

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Sum

If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`

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Solution

We have,

`LHS = m^2 – n^2 = (tanθ + sinθ)^2 – (tanθ – sinθ)^2`

`= 4tanθ sinθ [∵ (a + b)^2 – (a – b)^2 = 4ab]`

`=4\sqrt{(\tan \theta +\sin \theta )(\tan \theta \sin \theta )}`

`=4\sqrt{\tan ^{2}\theta \sin ^{2}\theta `

`=4\sqrt{\sin ^{2}\theta /\cos ^{2}\theta \sin^{2}\theta `

`=4\sqrt{(\sin ^{2}\theta \sin ^{2}\theta \cos^{2}\theta) /\cos^{2}\theta `

`=4\sqrt{(\sin ^{2}\theta (1\cos ^{2}\theta ))/\cos^{2}\theta }=4\sqrt{\sin ^{4}\theta /\cos ^{2}\theta `

`=\text{ }4\sin ^{2}\theta /\cos \theta =4\sin \theta\frac{\sin \theta }{\cos \theta }=4\sin \theta \tan \theta`

And, RHS = `4\sqrt{mn}`

Concept: Trigonometric Identities
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