sin 43° cos 47° + cos 43° sin 47° = ?
(a) 0
(b) 1
(c) sin 4°
(d) cos 4°

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Solution

$\sin 43 ° \cos 47 ° + \cos 43 ° \sin 47 ° = \sin (90 ° - 47 °) \cos 47 ° + \cos (90 ° - 47 °) \sin 47 ° = \cos 47 ° \cos 47 ° + \sin 47 ° \sin 47 ° (∵ \sin (90 ° - θ) = \cos θ and \cos (90 ° - θ) = sinθ) = \cos^{2} 47 ° + \sin^{2} 47 ° = 1 (using the identity : \sin^{2} θ + \cos^{2} θ = 1) Hence, the correct option is (b) .$

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Similar questions

sin 43^{\circ} cos 47^{\circ} + cos 43^{\circ} sin 47^{\circ} =

sin $47^{o}$ cos $43^{o}$ + cos $47^{o}$ sin $43^{o}$ = ?

(a) sin $4^{o}$ (b) cos $4^{o}$ (c) 1 (d) 0

{(\frac{\sin 47 °}{\cos 43 °})}^{2} + {(\frac{\cos 43 °}{\sin 47 °})}^{2} - 4 \cos^{2} 45 °

Simplify: ${(\frac{sin 47^{\circ}}{cos 43^{\circ}})}^{2} + {(\frac{cos 43^{\circ}}{sin 47^{\circ}})}^{2} - 4 {cos}^{2} 45^{\circ}$

s i n

43^{\circ}

\times

\frac{1}{c o s 47^{\circ}}

\times

s e c

63^{\circ}

\times

\frac{1}{c o s e c 27^{\circ}}

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