Statement-I: cosθ+isin θ5cos 3θ-i sin3θ6cos 2θ+isin2θ3cos 4θ-i sin 4θ5=1 Statement-II : cos2θ+isin 2θ3cos 3θ-i sin3θ4cos 3θ+isin3θ2cos 4θ+i sin 4θ-3=1

Statement-I:cosθ+isin θ5 cos 3θ-i sin3θ6cos 2θ+isin2θ3 cos 4θ-i sin 4θ5=1LHS=(cis θ)5 (cis(-3θ)6)(cis 2θ)3 (cis(-4θ))5 =cis5θ cis(-18θ)cis 6θ cis(-20θ) =cis (5θ-18θ-6θ+20θ) =cisθ∴ Statement I is false. Statement-II :LHS=cos2θ+isin 2θ3 cos 3θ-i sin3θ4cos 3θ+isin3θ2 cos 4θ+i sin 4θ-3 =(cis2θ)3 (cis(-3θ))4(cis3θ)2 (cis4θ)-3 =cis 6θ cis(-12θ)(cis 6θ) (cis(-12θ)) =1∴ Statement I is true.

Statement-I: cosθ+isin θ5cos 3θ-i sin3θ6cos 2θ+isin2θ3cos 4θ-i sin 4θ5=1 Statement-II : cos2θ+isin 2θ3cos 3θ-i sin3θ4cos 3θ+isin3θ2cos 4θ+i sin 4θ-3=1

Statement-I:cosθ+isin θ5 cos 3θ-i sin3θ6cos 2θ+isin2θ3 cos 4θ-i sin 4θ5=1LHS=(cis θ)5 (cis(-3θ)6)(cis 2θ)3 (cis(-4θ))5 =cis5θ cis(-18θ)cis 6θ cis(-20θ) =cis (5θ-18θ-6θ+20θ) =cisθ∴ Statement I is false. Statement-II :LHS=cos2θ+isin 2θ3 cos 3θ-i sin3θ4cos 3θ+isin3θ2 cos 4θ+i sin 4θ-3 =(cis2θ)3 (cis(-3θ))4(cis3θ)2 (cis4θ)-3 =cis 6θ cis(-12θ)(cis 6θ) (cis(-12θ)) =1∴ Statement I is true.

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$Statement - I : \frac{{(cosθ + isin θ)}^{5} {(\cos 3 θ - i \sin 3 θ)}^{6}}{{(\cos 2 θ + isin 2 θ)}^{3} {(\cos 4 θ - i \sin 4 θ)}^{5}} = 1 Statement - II : \frac{{(\cos 2 θ + isin 2 θ)}^{3} {(\cos 3 θ - i \sin 3 θ)}^{4}}{{(\cos 3 θ + isin 3 θ)}^{2} {(\cos 4 θ + i \sin 4 θ)}^{- 3}} = 1$

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$Only I is true$

$Only II is true$

$Both I and II are true$

$Neither I nor II are true$

answer is B.

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Detailed Solution

$\begin{array}{rcl} Statement - I : \\ \frac{{(cosθ + isin θ)}^{5} {(\cos 3 θ - i \sin 3 θ)}^{6}}{{(\cos 2 θ + isin 2 θ)}^{3} {(\cos 4 θ - i \sin 4 θ)}^{5}} & = & 1 \\ LHS & = & \frac{{(cis θ)}^{5} (cis {(- 3 θ)}^{6})}{{(cis 2 θ)}^{3} (cis {(- 4 θ))}^{5}} \\ = & \frac{cis 5 θ cis (- 18 θ)}{cis 6 θ cis (- 20 θ)} \\ = & cis (5 θ - 18 θ - 6 θ + 20 θ) \\ = & cisθ \end{array}$

$∴$ Statement I is false.

$\begin{array}{lcl} Statement - II : \\ LHS & = & \frac{{(\cos 2 θ + isin 2 θ)}^{3} {(\cos 3 θ - i \sin 3 θ)}^{4}}{{(\cos 3 θ + isin 3 θ)}^{2} {(\cos 4 θ + i \sin 4 θ)}^{- 3}} \\ = & \frac{{(cis 2 θ)}^{3} (cis {(- 3 θ))}^{4}}{{(cis 3 θ)}^{2} {(cis 4 θ)}^{- 3}} \\ = & \frac{cis 6 θ cis (- 12 θ)}{(cis 6 θ) (cis (- 12 θ))} \\ = & 1 \end{array}$