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Miscellaneous Exercise 3 · Q129

Q.Show that cot⁡−113−tan⁡−113=cot⁡−134\cot^{-1}\dfrac{1}{3} - \tan^{-1}\dfrac{1}{3} = \cot^{-1}\dfrac{3}{4}.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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cot⁡−113=tan⁡−13\cot^{-1}\dfrac13=\tan^{-1}3 (property ii, since x=1/3>0x=1/3>0... more precisely cot⁡θ=1/3  ⟹  tan⁡θ=3\cot\theta=1/3\implies\tan\theta=3). So L.H.S. =tan⁡−13−tan⁡−113=\tan^{-1}3-\tan^{-1}\dfrac13. By the tan-difference property (xvi), =tan⁡−1(3−131+3⋅13)=tan⁡−1(8/32)=tan⁡−143=\tan^{-1}\left(\dfrac{3-\frac13}{1+3\cdot\frac13}\right)=\tan^{-1}\left(\dfrac{8/3}{2}\right)=\tan^{-1}\dfrac43. Converting back: $\tan^{-1}\dfrac43=\cot^{-1 …

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