Question

Let $y={\sin }^{-1}\left(\sin 8\right)-{\tan }^{-1}\left(\tan 10\right)+{\cos }^{-1}\left(\cos 12\right)-{\sec }^{-1}\left(\sec 9\right)+{\cot }^{-1}\left(\cot 6\right)-{\text{cosec}}^{–1}\left(\text{cosec}7\right).$ If $y$ simplifies to $a\pi +b,$ then the value of $a-b$ is

Answer 1

${\sin }^{-1}\left(\sin 8\right)={\sin }^{-1}\left(\sin \right(3\pi -8\left)\right)=3\pi -8$
${\tan }^{-1}\left(\tan 10\right)={\tan }^{-1}\left(\tan \right(10-3\pi \left)\right)=10-3\pi$
${\cos }^{-1}\left(\cos 12\right)={\cos }^{-1}\left(\cos \right(4\pi -12\left)\right)=4\pi -12$
${\sec }^{-1}\left(\sec 9\right)={\sec }^{-1}\left(\sec \right(9-2\pi \left)\right)=9-2\pi$
${\cot }^{-1}\left(\cot 6\right)={\cot }^{-1}\left(\cot \right(6-\pi \left)\right)=6-\pi$
${\text{cosec}}^{-1}\left(\text{cosec}7\right)={\text{cosec}}^{-1}\left(\text{cosec}\right(7-2\pi \left)\right)=7-2\pi$

Now, $y=(3\pi -8)+(3\pi -10)+(4\pi -12)+(2\pi -9)+(-\pi +6)+(2\pi -7)$
$\Rightarrow y=13\pi -40=a\pi +b$
$\Rightarrow a=13$ and $b=-40$
$\therefore a-b=13+40=53$