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Q.
Column I Column II
A $ \cot ^{-1}\left(\tan \left(-37^{\circ}\right)\right)$ P $143^{\circ}$
B $ \cos ^{-1}\left(\cos \left(-233^{\circ}\right)\right)$ Q $127^{\circ}$
C $ \sin \left(\frac{1}{2} \cos ^{-1}\left(\frac{1}{9}\right)\right)$ R $\frac{3}{4}$
D $ \cos \left(\frac{1}{2} \arccos \left(\frac{1}{8}\right)\right)$ S $\frac{2}{3}$

Inverse Trigonometric Functions

Solution:

(A) $ \cot ^{-1}\left(\tan \left(-37^{\circ}\right)\right)=\cot ^{-1}\left(\cot \left(90+37^{\circ}\right)\right)=\cot ^{-1}\left(\cot \left(127^{\circ}\right)\right)=127^{\circ} \Rightarrow$ (Q)
(B) $ \cos ^{-1}\left(\cos \left(233^{\circ}\right)\right)=\cos ^{-1}\left(\cos \left(360^{\circ}-127^{\circ}\right)\right)=\cos ^{-1}\left(\cos \left(127^{\circ}\right)\right)=127^{\circ} \Rightarrow$ (Q)
(C) $\sin \frac{\theta}{2}$ where $\cos \theta=\frac{1}{9} ; 1-2 \sin ^2 \frac{\theta}{2}=\frac{1}{9}$
$\therefore 2 \sin ^2 \frac{\theta}{2}=1-\frac{1}{9}=\frac{8}{9} \Rightarrow \sin ^2 \frac{\theta}{2}=\frac{4}{9} \Rightarrow \sin \frac{\theta}{2}=\frac{2}{3} \Rightarrow$
(D) $\cos \frac{\theta}{2}$ where $\cos \theta=\frac{1}{8}$
$2 \cos ^2 \frac{\theta}{2}-1=\frac{1}{8} \Rightarrow 2 \cos ^2 \frac{\theta}{2}=1+\frac{1}{8}=\frac{9}{8} \Rightarrow \cos ^2 \frac{\theta}{2}=\frac{9}{16} $
$\cos \frac{\theta}{2}=\frac{3}{4} \Rightarrow \text { (R) }$