Question

If $A={\sin }^{-1}\left(\sin 10\right),B={\cos }^{-1}\left(\cos 10\right)$ then

• A. $A=3\pi -10$
• B. $A=3\pi +10$
• C. A > B
• D. A < B

Answer 1

Answer: (A), (D)

$A=3\pi -10$
$A<B$

Given:

$A={\sin }^{-1}\left(\sin 10\right)$
As $3\pi <10<3\pi +\frac{\pi }{2}$

[Since, $1$ radian $={57.29}^{\circ }$ (Approximately), $10$ radias $={572.9}^{\circ }$(Approximately),]
$A={\sin }^{-1}\left(\sin (3\pi -10)\right)=3\pi -10$

$B={\cos }^{-1}\left(\cos (10-3\pi )\right)=10-3\pi$

Since, $3\pi -10<10-3\pi$
Hence, $A<B$