Find the value of

sin A. cos B – cos A. sin B, if tan A= √3 and sin B = 1/2.

Given: tan A =√3 and

We know that,

Or

Let,

Side opposite to angle θ = 1k

and Hypotenuse = 2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AC)² + (BC)² = (AB)²

⇒ (1k)² + (BC)² = (2k)²

⇒ k² + (BC)² = 4k²

⇒ (BC)² = 4k² –k²

⇒ (BC)² = 3 k²

⇒ BC =√3k²

⇒ BC =k√3

So, BC = k√3

Now, we have to find the value of cos B

We know that,

The side adjacent to angle B = BC =k√3

Hypotenuse = AB =2k

So,

We know that,

Or

Given: tan A = √3

Let,

The side opposite to angle A =BC = √3k

The side adjacent to angle A =AB = 1k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (1k)² + (√3k)² = (AC)²

⇒ (AC)² = 1 k² +3 k²

⇒ (AC)² = 4 k²

⇒ AC =√2 k²

⇒ AC =±2k

But side AC can’t be negative. So, AC = 2k

Now, we will find the sin A and cos A

Side opposite to angle A = BC = k√3

and Hypotenuse = AC = 2k

So,

Now, we know that,

The side adjacent to angle A = AB =1k

Hypotenuse = AC =2k

So,

Now, sin A. cos B – cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get