Question

$\begin{array}{cccccc}& ColumnI& & Column2& & Column3\\ \left(I\right)& cos\left(si{n}^{-1}\right(sin\frac{5\pi }{6}\left)\right)=& \left(i\right)& 1& \left(P\right)& f\left(x\right)=3x^3-13x^2+14x-2has\\ & & & & & threeroots\alpha ,\beta ,\gamma ,thenvalueof\\ & & & & & |sin(ta{n}^{-1}\alpha +ta{n}^{-1}\beta +ta{n}^{-1}\gamma )|=\\ \left(II\right)& cos\left(co{s}^{-1}\right(-sin\frac{7\pi }{6}\left)\right)=& \left(ii\right)& \frac{1}{\sqrt{2}}& \left(Q\right)& \frac{(\sqrt{3}cosec{20}^{\circ }-sec{20}^{\circ })}{8}=\\ \left(III\right)& \left|cos\left\{ta{n}^{-1}\right(tan\frac{15\pi }{4}\left)\right\}\right|=& \left(iii\right)& \frac{1}{2}& \left(R\right)& 4\sqrt{2}\left(sin{12}^{\circ }\right)\left(sin{48}^{\circ }\right)\left(sin{54}^{\circ }\right)=\\ \left(IV\right)& sin({(co{s}^{-1}\left\{\frac{1}{\sqrt{2}}(cos\frac{9\pi }{10}-sin\frac{9\pi }{10})\right\}+\frac{7\pi }{20})}^{-1}\times {\pi }^2)& \left(iv\right)& \frac{\sqrt{3}}{2}& \left(S\right)& IftheanglesA,BandCofa\\ & & & & & triangleareinanarithmeticprogression\\ & & & & & andifa,bandcdenotethelengthsof\\ & & & & & sidesoppositetoA,BandCrespectively,\\ & & & & & thenthevalueoftheexpression\\ & & & & & \frac{1}{2}(\frac{a}{c}sin2C+\frac{c}{a}sin2A)is\end{array}$

Which of the following options is only correct combination?

• A. (I), (iv), (S)
• B. (B) (I), (iv), (R)
• C. (I), (iii), (S)
• D. (I), (iii), (i)

Answer 1

The correct option is A

(I), (iv), (S)

$cos\left(si{n}^{-1}\left(sin\frac{5\pi }{6}\right)\right)=cos\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}cos\left(co{s}^{-1}(-sin\frac{7\pi }{6})\right)=cos\left(\frac{\pi }{3}\right)=\frac{1}{2}cos\left(ta{n}^{-1}\left(tan\frac{15\pi }{4}\right)\right)=cos\left(\frac{3\pi }{4}\right)=\frac{-1}{\sqrt{2}}co{s}^{-1}\left\{\frac{1}{\sqrt{2}}(cos\frac{9\pi }{10}-sin\frac{9\pi }{10}-sin\frac{9\pi }{10})\right\}=co{s}^{-1}\left(cos\frac{23\pi }{20}\right)=2\pi -\frac{23\pi }{20}=\frac{17\pi }{20}$
$f\left(x\right)=3x^3-13x^2+14x-2\alpha ,\beta ,\gamma \alpha +\beta +\gamma =\frac{13}{3}\alpha \beta +\beta \gamma +\gamma \alpha =\frac{14}{3}\alpha \beta \gamma =\frac{2}{3}\therefore sin(ta{n}^{-1}\alpha +ta{n}^{-1}\beta +ta{n}^{-1}\gamma )sin\left(ta{n}^{-1}\left(\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}\right)\right)=sin\left(ta{n}^{-1}\right(-1\left)\right)=-\frac{1}{\sqrt{2}}$
$\sqrt{3}cosec{20}^{\circ }-sec{20}^{\circ }=tan{60}^{\circ }cosec{20}^{\circ }-sec{20}^{\circ }=\frac{sin{60}^{\circ }cos{20}^{\circ }-cos{60}^{\circ }sin{20}^{\circ }}{cos{60}^{\circ }.sin{20}^{\circ }cos{20}^{\circ }}=\frac{sin{40}^{\circ }}{\frac{1}{2}sin{20}^{\circ }cos{20}^{\circ }}=4\left(sin{12}^{\circ }\right)\left(sin{48}^{\circ }\right)\left(sin{54}^{\circ }\right)=\frac{1}{2}\left(2sin{12}^{\circ }sin{48}^{\circ }\right)sin{54}^{\circ }=\frac{1}{2}\left(2sin{12}^{\circ }sin{48}^{\circ }\right)sin{54}^{\circ }=\frac{1}{2}[cos{36}^{\circ }-cos{60}^{\circ }]sin{54}^{\circ }=\frac{1}{2}[cos{36}^{\circ }-\frac{1}{2}]sin{54}^{\circ }=\frac{1}{2}[{\left(\frac{\sqrt{5}+1}{4}\right)}^2-\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)]=\frac{1}{8}$
(iv)Since A,B,C are in A.P. $\angle B={60}^{\circ }$
$\therefore \frac{a}{c}\left(2sinCcosC\right)+\frac{c}{a}\left(2sinAcosA\right)=2k(acosC+ccosA)=2kb=2sinB=\sqrt{3}$