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  • If <img alt="\cos^{-1}\frac{4x}{1+4x^2}=-\frac{\pi}{2}+2tan^{-1}2x" src="https://learn.careers360.com/latex-image/?%5Ccos%5E%7B-1%7D%5Cfrac%7B4x%7D%7B1&plus;4x%5E2%7D%3D-%5Cfrac%7B%5Cpi%7D

If \cos^{-1}\frac{4x}{1+4x^2}=-\frac{\pi}{2}+2tan^{-1}2x then the value of x?

Option: 1

0


Option: 2

\frac{1}{4}


Option: 3

\frac{1}{3}


Option: 4

\frac{1}{2}


Answers (1)

best_answer

 

 

Multiple angles in terms of  arctan and arcsin -

Multiple angles in terms of  arctan and arcsin 

 

 

\\\mathrm{\;\;\;2\;\tan^{-1}x=\left\{\begin{matrix} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), & & \text { if }-1 \leq x \leq 1\\ \\ \pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right),& &\text { if }\;\;x>1 \\\\ -\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right),& & \text { if }\;\;x<-1 \end{matrix}\right.}


 

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\cos^{-1}\frac{4x}{1+4x^2}=-\frac{\pi}{2}+2tan^{-1}2x\\ \frac{\pi}{2}-\sin^{-1}\frac{4x}{1+4x^2}=-\frac{\pi}{2}+2tan^{-1}2x\\ \frac{\pi}{2}-\sin^{-1}\frac{4x}{1+(2x)^2}=-\frac{\pi}{2}+2tan^{-1}2x\\ \frac{\pi}{2}-2tan^{-1}2x=-\frac{\pi}{2}+2tan^{-1}2x\ \because \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2tan^{-1}x,-1 \leq x \leq 1\\ tan^{-1}2x=\frac{\pi}{4}\\ x=\frac{1}{2}

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