# If A = B = 60°, verify that

Question:

If $A=B=60^{\circ}$, verify that

(i) $\cos (A-B)=\cos A \cos B+\sin A \sin B$

(ii) $\sin (A-B)=\sin A \cos B-\cos A \sin B$

(iii) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

Solution:

(i) Given:

$A=B=60^{\circ} \ldots \ldots(1)$

To verify:

$\cos (A-B)=\cos A \cos B+\sin A \sin B$....(2)

Now consider left hand side of the expression to be verified in equation (2)

Therefore,

$\cos (A-B)=\cos (60-60)$

$=\cos 0$

$=1$

Now consider right hand side of the expression to be verified in equation (2)

Therefore,

$\cos A \cos B+\sin A \sin B=\cos B \cos B+\sin B \sin B$

$=\cos ^{2} B+\sin ^{2} B$

$=1$

Hence it is verified that,

$\cos (A-B)=\cos A \cos B+\sin A \sin B$

(ii) Given:

$A=B=60^{\circ} \ldots \ldots(1)$

To verify:

$\sin (A-B)=\sin A \cos B-\cos A \sin B$...(2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

$\sin (A-B)=\sin (B-B)$

$=\sin 0$

$=0$

Now by substituting the value of A and B from equation (1) in the above expression

We get,

$\sin A \cos B-\cos A \sin B=\sin B \cos B-\cos B \sin B$

= 0

Hence it is verified that,

$\sin (A-B)=\sin A \cos B-\cos A \sin B$

(iii) Given:

$A=B=60^{\circ}$...(1)

To verify:

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$....(2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

$\tan (A-B)=\tan (B-B)$

$=\tan 0$

$=0$

Now consider RHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of $A$ and $B$ from equation (1) in the above expression We get,

$\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{\tan B-\tan B}{1+\tan B \tan B}$

$=\frac{0}{1+\tan ^{2} B}$

$=0$

Hence it is verified that,

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$