Inverse Trigonometric Functions Question 96
Question: The value of $ {{\tan }^{-1}}( \frac{1}{2}\tan 2A)+{{\tan }^{-1}}(cotA)+ta{n^{-1}}(cot^{3}A) ) $ is
Options:
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
Show Answer
Answer:
Correct Answer: 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}}(cot^{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 ec^{2}A.{{\sin }^{4}}A}{{{\sin }^{4}}A-{{\cos }^{4}}A} $
$ =\frac{\sin A\cos A}{(sin^{2}A+{{\cos }^{2}}A)(sin^{2}A-{{\cos }^{2}}A)} $
$ =-\frac{\sin 2A}{2\cos 2A}=-\frac{1}{2}\tan 2A $
Hence, $ {{\tan }^{-1}}( \frac{1}{2}\tan 2A )+{{\tan }^{-1}}(CotA)+ta{n^{-1}}(cot^{3}A)=\pi , $
$ = \begin{matrix} \pi if0<A<\frac{\pi }{4} \\ 0if\frac{\pi }{4}<A<\frac{\pi }{2} \\ \end{matrix} . $ [Since, $ {{\tan }^{-1}}(-x)=-ta{n^{-1}}x $ ]
BETA
