Question

$\huge{prove\: that}$
$sin 70° cos 10° - cos 70° sin 10° = \frac{ \sqrt{3} }{2} \ \textless \ br /\ \textgreater$
​

Answer 1

LHS

sin 70°.cos 10° - cos 70° .sin 10°

We know that,

sin (A-B) = sin A. Cos B- CosA.sinB

So,

=> sin (70°-10°)

=> sin 60

RHS

$= \frac{ \sqrt{3} }{2}$

Hope it helps:)

Answer 2

Answer:

LHS

sin 70°.cos 10° - cos 70° .sin 10°

We know that,

sin (A-B) = sin A. Cos B- CosA.sinB

So,

=> sin (70°-10°)

=> sin 60

RHS

hope it yelps