Question

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	$\frac{2}{3}$

• A. $A\to P;B\to Q;C\to S;D\to R$
• B. $A\to Q;B\to Q;C\to S;D\to R$
• C. $A\to R;B\to Q;C\to S;D\to R$
• D. $A\to S;B\to Q;C\to S;D\to R$

Answer 1

Answer: (B)

(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 $\cos \theta =\frac{1}{8}$
$2{\cos }^2\frac{\theta }{2}-1=\frac{1}{8}\Rightarrow 2{\cos }^2\frac{\theta }{2}=1+\frac{1}{8}=\frac{9}{8}\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{9}{16}$
$\cos \frac{\theta }{2}=\frac{3}{4}\Rightarrow \text{ (R) }$