Q. 42 B3.7( 3 Votes )

Find the value of

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

Answer :

Given: tan A =√3 and

We know that,



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 =k3

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


We know that,


Given: tan A = √3


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


Now, we know that,

The side adjacent to angle A = AB =1k

Hypotenuse = AC =2k


Now, sin A. cos B – cos A. sin B

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

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