prove that tan-1(1/2 tan2A) + tan-1(cotA) + tan -1(cot3A) ={ 0, if pi/4 <A< pi/2 pi , if 0 < A < pi/4}

We have to prove:tan-112tan2A+tan-1cotA+tan-1cot3A=0,if π4 1Now, for 0 1 and for π4<A<π2; cotA<1Therefore, we can writetan-1cotA+tan-1cot3A=tan-1cotA+cot3A1-cot4A,if π4<A<π2π+tan-1cotA+cot3A1-cot4Aif 0<A<π2 =tan-1cotA1-cot2Aif π4<A<π2π+tan-1cotA1-cot2Aif 0<A<π2 As, cotA+cot3A1-cot4A=cotA(1+cot2A)(1-cot2A)(1+cot2A) Now, cotA1-cot2A=1tanA1-1tan2A= 1tanAtan2A-1tan2A=tanAtan2A-1=- tanA1-tan2A=-12tan2ATherefore, tan-1cotA+tan-1cot3A= tan-1-12tan2Aif π4<A<π2π+tan-1-12tan2Aif 0<A<π2 = -tan-112tan2Aif π4<A<π2π-tan-112tan2Aif 0<A<π2 Adding tan-112tan2A on both sides, we get tan-112tan2A+ tan-1cotA+tan-1cot3A= 0,if π4<A<π2π,if 0<A<π2

prove that tan^-1(1/2 tan2A) + tan^-1(cotA) + tan ^-1(cot³A) ={ 0, if pi/4 <A< pi/2
pi , if 0 < A < pi/4}

We have to prove : \tan^{- 1} (\frac{1}{2} \tan 2 A) + \tan^{- 1} (cotA) + \tan^{- 1} (\cot^{3} A) = \{\begin{cases} 0, & if \frac{π}{4} < A < \frac{π}{2} \\ π, & if 0 < A < \frac{π}{4} \end{cases} We know that \tan^{- 1} x + \tan^{- 1} y = \{\begin{cases} \tan^{- 1} (\frac{x + y}{1 - xy}), & if xy < 1 \\ π + \tan^{- 1} (\frac{x + y}{1 - xy}), & if xy > 1 \end{cases} Now, for 0 < A < \frac{π}{4}; cotA > 1 and for \frac{π}{4} < A < \frac{π}{2}; cotA < 1 Therefore, we can write \tan^{- 1} (cotA) + \tan^{- 1} (\cot^{3} A) = \{\begin{cases} \tan^{- 1} (\frac{cotA + \cot^{3} A}{1 - \cot^{4} A}), & if \frac{π}{4} < A < \frac{π}{2} \\ π + \tan^{- 1} (\frac{cotA + \cot^{3} A}{1 - \cot^{4} A}) & if 0 < A < \frac{π}{2} \end{cases} = \{\begin{cases} \tan^{- 1} (\frac{cotA}{1 - \cot^{2} A}) & if \frac{π}{4} < A < \frac{π}{2} \\ π + \tan^{- 1} (\frac{cotA}{1 - \cot^{2} A}) & if 0 < A < \frac{π}{2} \end{cases} [As, \frac{cotA + \cot^{3} A}{1 - \cot^{4} A} = \frac{cotA (1 + \cot^{2} A)}{(1 - \cot^{2} A) (1 + \cot^{2} A)}] Now, \frac{cotA}{1 - \cot^{2} A} = \frac{\frac{1}{tanA}}{1 - \frac{1}{\tan^{2} A}} = \frac{\frac{1}{tanA}}{\frac{\tan^{2} A - 1}{\tan^{2} A}} = \frac{tanA}{\tan^{2} A - 1} = - \frac{tanA}{1 - \tan^{2} A} = - \frac{1}{2} \tan 2 A Therefore, \tan^{- 1} (cotA) + \tan^{- 1} (\cot^{3} A) = \{\begin{cases} \tan^{- 1} (- \frac{1}{2} \tan 2 A) & if \frac{π}{4} < A < \frac{π}{2} \\ π + \tan^{- 1} (- \frac{1}{2} \tan 2 A) & if 0 < A < \frac{π}{2} \end{cases} = \{\begin{cases} - \tan^{- 1} (\frac{1}{2} \tan 2 A) & if \frac{π}{4} < A < \frac{π}{2} \\ π - \tan^{- 1} (\frac{1}{2} \tan 2 A) & if 0 < A < \frac{π}{2} \end{cases} Adding \tan^{- 1} (\frac{1}{2} \tan 2 A) on both sides, we get \tan^{- 1} (\frac{1}{2} \tan 2 A) + \tan^{- 1} (cotA) + \tan^{- 1} (\cot^{3} A) = \{\begin{cases} 0, & if \frac{π}{4} < A < \frac{π}{2} \\ π, & if 0 < A < \frac{π}{2} \end{cases}

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