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3.Trigonometrical Ratios, Functions and Identities
medium
If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n,$ then
A
${m^2} - {n^2} = 4\,mn$
B
${m^2} + {n^2} = 4\,mn$
C
${m^2} - {n^2} = {m^2} + {n^2}$
D
${m^2} - {n^2} = 4\sqrt {mn} $
(IIT-1970)
Solution
(d) $(m + n) = 2\,\tan \theta ,\,\,m – n = 2\,\sin \theta $
$\therefore \,\,\,{m^2} – {n^2} = 4\,\tan \theta \,.\,\sin \theta $…..$(i)$
$4\sqrt {mn} = 4\sqrt {{{\tan }^2}\theta – {{\sin }^2}\theta } $
$= 4\,\sin \theta \,.\,\tan \theta $…..$(ii)$
From $(i)$ and $(ii)$, ${m^2} – {n^2} = 4\sqrt {mn} $.
Standard 11
Mathematics
