Question

Sin 70 degree×cos 10 degree-cos70× sin 10 = root3/2

Answer 1

answer of your question................

Answer 2

Answer:

The correct answer is,

$sin70^0cos 10^0-cos70^0sin10^0$ $=\frac{\sqrt{3} }{2}$

Step-by-step explanation:

To Prove:

$sin70^0cos 10^0-cos70^0sin10^0$ $=\frac{\sqrt{3} }{2}$

We know,

sin(a-b)=sin a cos b - cos a sin b

Therefore,

⇒ $sin70^0cos 10^0-cos70^0sin10^0$

   $= sin(70-10)$

⇒ $sin70^0cos 10^0-cos70^0sin10^0$

   $= sin 60$

⇒ $sin70^0cos 10^0-cos70^0sin10^0$

    $=\frac{\sqrt{3} }{2}$

L.H.S. = R.H.S.

Hence proved.

#SPJ3