If tan theta + sin theta = m and tan theta - | vedclass

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(D) Given: $m = 	an 	heta + \sin 	heta$ and $n = 	an 	heta - \sin 	heta$.First,calculate $m^2 - n^2$:$m^2 - n^2 = (m + n)(m - n)$$m + n = (	an

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${m^2} - {n^2} = 4\sqrt {mn} $

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(D) Given: $m = 	an 	heta + \sin 	heta$ and $n = 	an 	heta - \sin 	heta$.First,calculate $m^2 - n^2$:$m^2 - n^2 = (m + n)(m - n)$$m + n = (	an 	heta + \sin 	heta) + (	an 	heta - \sin 	heta) = 2 	an 	heta$$m - n = (	an 	heta + \sin 	heta) - (	an 	heta - \sin 	heta) = 2 \sin 	heta$Therefore,$m^2 - n^2 = (2 	an 	heta)(2 \sin 	heta) = 4 	an 	heta \sin 	heta$ ..... $(i)$Next,calculate $4\sqrt{mn}$:$mn = (	an 	heta + \sin 	heta)(	an 	heta - \sin 	heta) = 	an^2 	heta - \sin^2 	heta$$mn = \frac{\sin^2 	heta}{\cos^2 	heta} - \sin^2 	heta = \sin^2 	heta \left( \frac{1}{\cos^2 	heta} - 1 ight) = \sin^2 	heta \left( \frac{1 - \cos^2 	heta}{\cos^2 	heta} ight) = \sin^2 	heta \left( \frac{\sin^2 	heta}{\cos^2 	heta} ight) = \sin^2 	heta 	an^2 	heta$So,$4\sqrt{mn} = 4\sqrt{\sin^2 	heta 	an^2 	heta} = 4 \sin 	heta 	an 	heta$ ..... $(ii)$From $(i)$ and $(ii)$,we get $m^2 - n^2 = 4\sqrt{mn}$.

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n,$ then

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