Match the following from List – I to List – II LIST – I LIST – II(A)sin(410°−A)cos(400°+A)+cos(410°−A)sin(400°+A)has the value equal to (I)−1(B)cos21°−cos22°2sin3°sin1° is equal to(II)0(C)sin(−870°)+cosec(−660°)+tan(−855°)+2cot(840°)+cos(480°)+sec(900°)=(III)12(D)If cosθ=45where θ∈(3π2, 2π)and cosϕ=35where ϕ∈(0, π2) then cos(θ−ϕ) has the value equal to(IV)1 (V)2

A) sin(410°+400°)=sin810°=sin(720°+90°)=sin90°=1⇒SB) sin22°−sin21°2sin3°sin1°=sin3°sin1°2sin3°sin1°=12⇒(R)C) −sin(810°+60°)−cosec(720°−60°)−tan(−810°+45°)+2cos(810°+45°)+2cos120°+cos120°+sec180°=−12+23+1−23−12−1=−1⇒(P)D) cos(θ−ϕ)=cosθcosϕ+sinθsinϕ=45×35−35×45=0⇒(Q)

Match the following from List – I to List – II LIST – I LIST – II(A)sin(410°−A)cos(400°+A)+cos(410°−A)sin(400°+A)has the value equal to (I)−1(B)cos21°−cos22°2sin3°sin1° is equal to(II)0(C)sin(−870°)+cosec(−660°)+tan(−855°)+2cot(840°)+cos(480°)+sec(900°)=(III)12(D)If cosθ=45where θ∈(3π2, 2π)and cosϕ=35where ϕ∈(0, π2) then cos(θ−ϕ) has the value equal to(IV)1 (V)2

A) sin(410°+400°)=sin810°=sin(720°+90°)=sin90°=1⇒SB) sin22°−sin21°2sin3°sin1°=sin3°sin1°2sin3°sin1°=12⇒(R)C) −sin(810°+60°)−cosec(720°−60°)−tan(−810°+45°)+2cos(810°+45°)+2cos120°+cos120°+sec180°=−12+23+1−23−12−1=−1⇒(P)D) cos(θ−ϕ)=cosθcosϕ+sinθsinϕ=45×35−35×45=0⇒(Q)

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Match the following from List – I to List – II
LIST – I LIST – II
(A)
$\sin (410 ° - A) \cos (400 ° + A) + \cos (410 ° - A) \sin (400 ° + A)$
has the value equal to
(I) $- 1$
(B) $\frac{\cos^{2} 1 ° - \cos^{2} 2 °}{2 \sin 3 ° \sin 1 °}$ is equal to (II) 0
(C) $\begin{array}{l} \sin (- 870 °) + \cos e c (- 660 °) + \tan (- 855 °) + 2 \cot (840 °) \\ + \cos (480 °) + \sec (900 °) = \end{array}$ (III) $\frac{1}{2}$
(D)
If $\cos θ = \frac{4}{5}$ where $θ \in (\frac{3 π}{2}, 2 π)$ and $\cos ϕ = \frac{3}{5}$ where
$ϕ \in (0, \frac{π}{2})$ then $\cos (θ - ϕ)$ has the value equal to
(IV)
1

(V) 2

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$A - I V; B - I I I; C - I; D - I I$

$A - I I I; B - I V; C - I; D - I I$

$A - I V; B - I I I; C - V; D - I$

$A - I V; B - I; C - I I I; D - V$

answer is A.

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Detailed Solution

A) $\sin (410 ° + 400 °) = \sin 810 ° = \sin (720 ° + 90 °) = \sin 90 ° = 1 \Rightarrow S$

B) $\frac{\sin^{2} 2 ° - \sin^{2} 1 °}{2 \sin 3 ° \sin 1 °} = \frac{\sin 3 ° \sin 1 °}{2 \sin 3 ° \sin 1 °} = \frac{1}{2} \Rightarrow (R)$

C) $\begin{array}{l} - \sin (810 ° + 60 °) - \cos e c (720 ° - 60 °) - \tan (- 810 ° + 45 °) + 2 \cos (810 ° + 45 °) \\ + 2 \cos 120 ° + \cos 120 ° + \sec 180 ° = - \frac{1}{2} + \frac{2}{\sqrt{3}} + 1 - \frac{2}{\sqrt{3}} - \frac{1}{2} - 1 = - 1 \Rightarrow (P) \end{array}$

D) $\cos (θ - ϕ) = \cos θ \cos ϕ + \sin θ \sin ϕ = \frac{4}{5} \times \frac{3}{5} - \frac{3}{5} \times \frac{4}{5} = 0 \Rightarrow (Q)$

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