Question

$3ta{n}^{-1}\left(\frac{1}{2}\right)+2ta{n}^{-1}\left(\frac{1}{5}\right)+si{n}^{-1}\left(\frac{142}{65\sqrt{5}}\right)=$

• A. $0$
• B. $\pi$
• C. $\frac{3\pi }{2}$
• D. $2\pi$

Answer 1

The correct option is B $\pi$

$3ta{n}^{-1}\frac{1}{2}+2ta{n}^{-1}\frac{1}{5}+si{n}^{-1}\frac{142}{65\sqrt{5}}=2[ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{5}]+ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{142}{31}=2ta{n}^{-1}\frac{7}{9}+\pi +ta{n}^{-1}\frac{\frac{1}{2}+\frac{142}{31}}{1-\frac{1}{2}\times \frac{142}{31}}[\because \frac{1}{2}\times \frac{142}{31}>1]=ta{n}^{-1}\frac{2\times \frac{7}{9}}{1-\frac{49}{31}}+\pi -ta{n}^{-1}\frac{315}{80}=\pi +ta{n}^{-1}\frac{63}{16}-ta{n}^{-1}\frac{63}{16}=\pi$