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  • If <img alt="\sin^{-1}(\frac{8x}{x^2+16})+2\tan^{-1}(\frac{-x}{4})" src="https://learn.careers360.com/latex-image/?%5Csin%5E%7B-1%7D%28%5Cfrac%7B8x%7D%7Bx%5E2&plus;16%7D%29&plus;2%5Cta

If \sin^{-1}(\frac{8x}{x^2+16})+2\tan^{-1}(\frac{-x}{4}) is independent of x then which of the following is true?

Option: 1

x\epsilon [-2,2]


Option: 2

x\epsilon [-\infty,\infty]


Option: 3

x\epsilon [0,\infty]


Option: 4

x\epsilon [-4,4]


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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\sin^{-1}(\frac{8x}{x^2+16})+2\tan^{-1}(\frac{-x}{4})=\sin ^{-1}(\frac{2 \cdot \frac{x}{4}}{(\frac{x}{4})^2+1})-2\tan^{-1}(\frac{x}{4})\\ =2\tan^{-1}(\frac{x}{4})-2\tan^{-1}(\frac{x}{4})\ \because \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2\tan^{-1}x, \text { if }-1 \leq x \leq 1\\ =0\\ \text{so } -1\leq\frac{x}{4} \leq 1\\ x\epsilon [-4,4]

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Gaurav

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