Question

The value of $3{\tan }^{-1}\frac{1}{2}+2{\tan }^{-1}\frac{1}{5}+{\sin }^{-1}\frac{142}{65\sqrt{5}}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $noneofthese$

Answer 1

Answer: (C)

$\pi$

$\Rightarrow 3{\tan }^{-1}\frac{1}{2}+2{\tan }^{-1}\frac{1}{5}+{\sin }^{-1}\frac{142}{65\sqrt{5}}$

$\Rightarrow ({\tan }^{-1}\frac{1}{\alpha }+{\tan }^{-1}\frac{1}{\alpha })+{\tan }^{-1}\frac{1}{\alpha }+({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{5})+{\tan }^{-1}\left(\frac{142}{31}\right)$

$\Rightarrow ({\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{\alpha })+{\tan }^{-1}\left(\frac{5}{12}\right)+{\tan }^{-1}\left(\frac{142}{31}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{11}{2}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)+{\tan }^{-1}\left(\frac{142}{31}\right)$

$\Rightarrow \pi +{\tan }^{-1}\left(\frac{-142}{31}\right)+{\tan }^{-1}\left(\frac{142}{31}\right)$

$\Rightarrow \pi$

Important property used

${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

for $x>o,y>o$ $xy<1$

${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

for $x>0,y>0,xy>1$.