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Solve 2tan-1(cosx) = tan-1(2cosecx).

Option: 1

n\pi


Option: 2

n\pi-\frac{\pi }{4}


Option: 3

n\pi+\frac{\pi }{4}


Option: 4

n\pi+\frac{\pi }{2}


Answers (1)

best_answer

L.H.S. =tan^{-1}\frac{{2\cos x}}{{1 - {{\cos }^2}x}} = {\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{{{\sin }^2}x}}} \right)        

                        = {\tan ^{ - 1}}\left[ {2\cot x\cos ecx} \right]

R.H.S.             = tan^{-1}(2 cosec x)          (given)

Hence, LHS = RHS

\Rightarrow {\tan ^{ - 1}}\left[ {2\cot x\cos ecx} \right] = \tan^{-1}[2cosecx]

                        \Rightarrow cotx = 1        \Rightarrow x = n\pi +\frac{\pi}{4}.

 

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