A) 0 if \[\frac{\pi }{4}<A<\frac{\pi }{2}\]
B) \[\pi \], if \[0<A<\frac{\pi }{4}\]
C) Both a and b
D) None of these
Correct Answer: C
Solution :
[c] We know that \[\cot A>1\] if \[0<A<\frac{\pi }{4}\] and \[\cot A<1if\frac{\pi }{4}<A<\frac{\pi }{2}\] \[{{\tan }^{-1}}(cotA)+ta{{n}^{-1}}(co{{t}^{3}}A)\] \[=\pi +{{\tan }^{-1}}\frac{\cot A+{{\cot }^{3}}A}{1-{{\cot }^{4}}A},\] If \[0<A<\frac{\pi }{4}\] and \[={{\tan }^{-1}}\frac{\cot A+{{\cot }^{3}}A}{1-{{\cot }^{4}}A}\] if \[\frac{\pi }{4}<A<\frac{\pi }{2}\] Also, \[\frac{\cot A+{{\cot }^{3}}A}{1-{{\cot }^{4}}A}=\frac{\cot A\cos e{{c}^{2}}A.{{\sin }^{4}}A}{{{\sin }^{4}}A-{{\cos }^{4}}A}\] \[=\frac{\sin A\cos A}{(si{{n}^{2}}A+{{\cos }^{2}}A)(si{{n}^{2}}A-{{\cos }^{2}}A)}\] \[=-\frac{\sin 2A}{2\cos 2A}=-\frac{1}{2}\tan 2A\] Hence, \[{{\tan }^{-1}}\left( \frac{1}{2}\tan 2A \right)+{{\tan }^{-1}}(CotA)+ta{{n}^{-1}}(co{{t}^{3}}A)=\pi ,\]\[=\left\{ \begin{matrix} \pi if0<A<\frac{\pi }{4} \\ 0if\frac{\pi }{4}<A<\frac{\pi }{2} \\ \end{matrix} \right.\] [Since,\[{{\tan }^{-1}}(-x)=-ta{{n}^{-1}}x\]]
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