Trig1.4

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If \tan\theta+\sin\theta=m,\tan\theta-\sin\theta=n\, show that m^2-n^2=4\sqrt{mn}\,

Given \tan\theta+\sin\theta=m,\tan\theta-\sin\theta=n\,

LHS= m^2-n^2=(\tan\theta+\sin\theta)^2-(\tan\theta-\sin\theta)^2\,

m^2-n^2=4\tan\theta\cdot\sin\theta=4\frac{\sin\theta}{\cos\theta}\cdot\sin\theta=4\frac{\sin^2 \theta}{\cos\theta}\,

m^2-n^2=4\sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}}=4\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}\cdot(1-\cos^2 \theta)}\,

m^2-n^2=4\sqrt{\tan^2 \theta-\sin^2 \theta}=4\sqrt{(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)}=4\sqrt{mn}\,=RHS


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