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,



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 =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


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






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