Question

prove that sin70°cos10°-cos70°sin10°=√3/2

Answer 1

Answer:To Prove: sin70^0cos 10^0–cos70^0sin10^0

Step-by-step explanation:

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 = √3/2

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

Hence proved.

Answer 2

Given,

A trigonometric equation  sin70°cos10°-cos70°sin10°=√3/2.

To Prove,

The value of  sin70°cos10°-cos70°sin10°=√3/2.

Solution,

Using the trigonometric formula

Sin(A-B) = sinA.cosB-cosA.sinB

Now, the given equation is

sin70°cos10°-cos70°sin10°=√3/2

Takin L.H.S

sin70°cos10°-cos70°sin10°

Now, using the formula of sin(A-B)

sin(70-10) = sin70°cos10°-cos70°sin10°

sin(60) = sin70°cos10°-cos70°sin10°

sin(60) = √3/2 = R.H.S

R.H.S = L.H.S

Hence proved

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