Question

The value of ${\sec }^{-1}\left(\sec 3\right)+{\cos }^{-1}\left(\cos 12\right)+{\text{cosec}}^{-1}\left(\text{cosec}6\right)+{\cot }^{-1}\left(\cot 10\right)$ is equal to

• A. $7-\pi$
• B. $\pi +7$
• C. $7$
• D. $\pi -7$

Answer 1

The correct option is A $7-\pi$

Let $y={\sec }^{-1}\left(\sec 3\right)+{\cos }^{-1}\left(\cos 12\right)+{\text{cosec}}^{-1}\left(\text{cosec}6\right)+{\cot }^{-1}\left(\cot 10\right),$
We know that
${\sec }^{-1}\left(\sec x\right)=x;\frac{\pi }{2}<x<\pi$
${\cos }^{-1}\left(\cos x\right)=4\pi -x;3\pi <x<4\pi$
${\text{cosec}}^{-1}\left(\text{cosec}x\right)=x-2\pi ;\frac{3\pi }{2}<x<2\pi$
${\cot }^{-1}\left(\cot x\right)=x-3\pi ;3\pi <x<4\pi$
$\Rightarrow y=3+4\pi -12+6-2\pi +10-3\pi$
$\therefore y=-\pi +7$