Question

$ta{n}^{-1}\frac{5}{6}+\frac{1}{2}ta{n}^{-1}\frac{11}{60}=$

• A. $\pi$
• B. $\pi /2$
• C. $\pi /4$
• D. $3\pi /4$

Answer 1

The correct option is C $\pi /4$

$ta{n}^{-1}\frac{5}{6}+\frac{1}{2}ta{n}^{-1}\frac{11}{60}$

$\frac{1}{2}ta{n}^{-1}\left(\frac{\frac{5}{6}\times 2}{1-(\frac{5}{6}{)}^2}\right)+\frac{1}{2}ta{n}^{-1}\left(\frac{11}{60}\right)[\because 2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}]$

$=\frac{1}{2}ta{n}^{-1}\left(\frac{60}{11}\right)+\frac{1}{2}ta{n}^{-1}\left(\frac{11}{60}\right)$

$=\frac{1}{2}\left[co{t}^{-1}\right(\frac{11}{60})+ta{n}^{-1}(\frac{11}{60}\left)\right][ta{n}^{-1}x=co{t}^{-1}\frac{1}{x}]$

$=\frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{4}$ Because$[co{t}^{-1}x+ta{n}^{-1}x=\frac{\pi }{2}]$