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

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Question

(d) $(m + n) = 2\,	an 	heta ,\,\,m - n = 2\,\sin 	heta $

$	herefore \,\,\,{m^2} - {n^2} = 4\,	an 	heta \,.\,\sin 	heta $&hellip;..$(i)$

$

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $(m + n) = 2\,	an 	heta ,\,\,m - n = 2\,\sin 	heta $

$	herefore \,\,\,{m^2} - {n^2} = 4\,	an 	heta \,.\,\sin 	heta $&hellip;..$(i)$

$4\sqrt {mn} = 4\sqrt {{{	an }^2}	heta - {{\sin }^2}	heta } $

$= 4\,\sin 	heta \,.\,	an 	heta $&hellip;..$(ii)$

From $(i)$ and $(ii)$, ${m^2} - {n^2} = 4\sqrt {mn} $.

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

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