Discussion Forum



1)

 If   $\alpha=3$   $\sin^{-1}\left(\frac{6}{11}\right)$   and    $\beta=3\cos^{-1}\left(\frac{4}{9}\right)$  , where the inverse trigonometric  functions take only the principal values, then the correct option (s) is /are


A) $\cos\beta>0$

B) $\sin\beta<0$

C) $\cos(\alpha+\beta)>0$

D) $\cos\alpha<0$

Answer:

Option A,C,D

Explanation:

 Here,  $\alpha=3\sin^{-}\left(\frac{6}{11}\right)$

    and    $\beta=3\cos^{-1}\left(\frac{4}{9}\right)4$   as   $\frac{6}{11}>\frac{1}{2}$

   $\Rightarrow $     $ \sin^{-1} \left(\frac{6}{11}\right)>\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

$\therefore$       $\alpha=3 \sin^{-1} \left(\frac{6}{11}\right)>\frac{\pi}{2}\Rightarrow\cos\alpha<0$

Now,    $\beta=3\cos^{-1}\left(\frac{4}{9}\right)$

As  $\frac{4}{9}<\frac{1}{2}\Rightarrow\cos^{-1}\left(\frac{4}{9}\right)>\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

$\therefore$        $\beta=3\cos^{-1}\left(\frac{4}{9}\right)>\pi$

$\therefore$                        $\cos\beta<0$    and    $\sin\beta<0$

Now, $\alpha+\beta$  is slightly greater than  $\frac{3\pi}{2}$

     $\therefore$          $\cos(\alpha+\beta)>0$