Question

The value of ${\sin }^{-1}\left(\sin 3\right)+{\cos }^{-1}\left(\cos 4\right)+{\tan }^{-1}\left(\tan 6\right)+{\cot }^{-1}\left(\cot 5\right)$ is equal to

• A. $10$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Let $y={\sin }^{-1}\left(\sin 3\right)+{\cos }^{-1}\left(\cos 4\right)+{\tan }^{-1}\left(\tan 6\right)+{\cot }^{-1}\left(\cot 5\right)$
We know that
${\sin }^{-1}\left(\sin x\right)=\pi -x;\frac{\pi }{2}<x<\frac{3\pi }{2}$
${\cos }^{-1}\left(\cos x\right)=2\pi -x;\pi <x<2\pi$
${\tan }^{-1}\left(\tan x\right)=x-2\pi ;\frac{3\pi }{2}<x<2\pi$
${\cot }^{-1}\left(\cot x\right)=x-\pi ;\pi <x<2\pi$
$\Rightarrow y=\pi -3+2\pi -4+6-2\pi +5-\pi$
$\therefore y=4$