Question

Which of the following is not the value of $2{\tan }^{-1}x$ ?

• A. ${\tan }^{-1}\frac{2x}{1+x^2}$
• B. ${\tan }^{-1}\frac{2x}{1-x^2}$
• C. ${\cos }^{-1}\frac{1-x^2}{1+x^2}$
• D. ${\sin }^{-1}\frac{2x}{1+x^2}$

Answer 1

Answer: (A)

Since $\frac{2tan\theta }{1-ta{n}^2\theta }=tan2\theta$
$\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=cos2\theta ,\frac{2tan\theta }{1+tan\theta }=sin2\theta$
$\therefore \left(b\right)=2\theta ,\left(c\right)=2\theta ,\left(d\right)=2\theta =2ta{n}^{-1}x$
Hence $ta{n}^{-1}\frac{2x}{1+x^2}\ne 2ta{n}^{-1}x$