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

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

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Solution

We know that, sin (A+B)= sin A cos B + cos A Sin B
Here a = $43^{o}$ and b = $47^{o}$

So, sin $47^{o}$ cos $43^{o}$ + cos $47^{o}$ sin $43^{o}$ = sin $(47^{o} + 43^{o})$

= sin $90^{o}$

= 1

Hence, the correct choice is (c).

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Similar questions

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

{(\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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