Matching Let cos(θ+700)=−13 where θ∈(00,1100) Column I Column IIPtan(θ+700)=122Qcos(1600+θ)=29+427Rsin(200−θ)=3−22Stan(250+θ)=4−223 5−13

cos(θ+700)=−13where θ∈(00,1100)P. θ+700∈(700,1800) tan(θ+700)=−22Q. cos(1600+θ)=cos(900+θ+700) = −sin(θ+700)=−223R. sin(200−θ)=sin[90−(θ+700)]=cos(θ+700)=−13S. 2n=1024 n=10 =tan(700+θ)−11+tan(700+θ)=9+427

Matching Let cos(θ+700)=−13 where θ∈(00,1100) Column I Column IIPtan(θ+700)=122Qcos(1600+θ)=29+427Rsin(200−θ)=3−22Stan(250+θ)=4−223 5−13

cos(θ+700)=−13where θ∈(00,1100)P. θ+700∈(700,1800) tan(θ+700)=−22Q. cos(1600+θ)=cos(900+θ+700) = −sin(θ+700)=−223R. sin(200−θ)=sin[90−(θ+700)]=cos(θ+700)=−13S. 2n=1024 n=10 =tan(700+θ)−11+tan(700+θ)=9+427

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Matching Let $\cos (θ + 70^{0}) = \frac{- 1}{3} w h e r e θ \in (0^{0}, 110^{0})$
Column I Column II
P $\tan (θ + 70^{0}) =$ 1 $2 \sqrt{2}$
Q $\cos (160^{0} + θ) =$ 2 $\frac{9 + 4 \sqrt{2}}{7}$
R $\sin (20^{0} - θ) =$ 3 $- 2 \sqrt{2}$
S $\tan (25^{0} + θ) =$ 4 $\frac{- 2 \sqrt{2}}{3}$
5 $\frac{- 1}{3}$

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$P \to 5, Q \to 3, R \to 4, S \to 1$

$P \to 3, Q \to 4, R \to 5, S \to 2$

$P \to 3, Q \to 5, R \to 2, S \to 4$

$P \to 1, Q \to 2, R \to 4, S \to 3$

answer is B.

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Detailed Solution

$\cos (θ + 70^{0}) = \frac{- 1}{3} w h e r e$ $θ \in (0^{0}, 110^{0})$

P. $θ + 70^{0} \in (70^{0}, 180^{0})$ $\tan (θ + 70^{0}) = - 2 \sqrt{2}$
Q. $\cos (160^{0} + θ) = \cos (90^{0} + θ + 70^{0})$ = $- \sin (θ + 70^{0}) = \frac{- 2 \sqrt{2}}{3}$
R. $\sin (20^{0} - θ) = \sin [90 - (θ + 70^{0})] = \cos (θ + 70^{0}) = \frac{- 1}{3}$
S. $2^{n} = 1024$ $n = 10$ $= \frac{\tan (70^{0} + θ) - 1}{1 + \tan (70^{0} + θ)} = \frac{9 + 4 \sqrt{2}}{7}$

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	Column I		Column II
P	$\tan (θ + 70^{0}) =$	1	$2 \sqrt{2}$
Q	$\cos (160^{0} + θ) =$	2	$\frac{9 + 4 \sqrt{2}}{7}$
R	$\sin (20^{0} - θ) =$	3	$- 2 \sqrt{2}$
S	$\tan (25^{0} + θ) =$	4	$\frac{- 2 \sqrt{2}}{3}$
		5	$\frac{- 1}{3}$

Matching Let cos(θ+700)=−13 where θ∈(00,1100) Column I Column IIPtan(θ+700)=122Qcos(1600+θ)=29+427Rsin(200−θ)=3−22Stan(250+θ)=4−223 5−13

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