Prove that:

Question:

Prove that:

(i) $\frac{\sin (A+B)+\sin (A-B)}{\cos (A+B)+\cos (A-B)}=\tan A$

(ii) $\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}=0$

(iii) $\frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A}=0$

(iv) $\sin ^{2} B=\sin ^{2} A+\sin ^{2}(A-B)-2 \sin A \cos B \sin (A-B)$

(v) $\cos ^{2} A+\cos ^{2} B-2 \cos A \cos B \cos (A+B)=\sin ^{2}(A+B)$

(vi) $\frac{\tan (A+B)}{\cot (A-B)}=\frac{\tan ^{2} A-\tan ^{2} B}{1-\tan ^{2} A \tan ^{2} B}$

Solution:

(i) LHS $=\frac{\sin (A+B)+\sin (A-B)}{\cos (A+B)+\cos (A-B)}$

$=\frac{\sin A \cos B+\cos A \sin B+\sin A \cos B-\cos A \sin B}{\cos A \cos B-\sin A \sin B+\cos A \cos B+\sin A \sin B}$

$=\frac{2 \sin A \cos B}{2 \cos A \cos B}$

$=\frac{\sin A}{\cos A}$

$=\tan A$

= RHS

Hence proved.

(ii) $\mathrm{LHS}=\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}$

$=\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B}+\frac{\sin B \cos C-\cos B \sin C}{\cos B \cos C}+\frac{\sin C \cos A-\cos C \sin A}{\cos C \cos A}$

$=\frac{\sin A \cos B}{\cos A \cos B}-\frac{\cos A \sin B}{\cos A \cos B}+\frac{\sin B \cos C}{\cos B \cos C}-\frac{\cos B \sin C}{\cos B \cos C}+\frac{\sin C \cos A}{\cos C \cos A}-\frac{\cos C \sin A}{\cos C \cos A}$

$=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}+\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}+\frac{\sin C}{\cos C}-\frac{\sin A}{\cos A}$

$=\tan A-\tan B+\tan B-\tan C+\tan C-\tan A$

$=0$

= RHS

Hence proved.

(iii) $\mathrm{LHS}=\frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A}$

$=\frac{\sin A \cos B-\cos A \sin B}{\sin A \sin B}+\frac{\sin B \cos C-\cos B \sin C}{\sin B \sin C}+\frac{\sin C \cos A-\cos C \sin A}{\sin C \sin A}$

$=\frac{\sin A \cos B}{\sin A \sin B}-\frac{\cos A \sin B}{\sin A \sin B}+\frac{\sin B \cos C}{\sin B \sin C}-\frac{\cos B \sin C}{\sin B \sin C}+\frac{\sin C \cos A}{\sin C \sin A}-\frac{\cos C \sin A}{\sin C \sin A}$

$=\frac{\cos B}{\sin B}-\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}-\frac{\cos B}{\sin B}+\frac{\cos A}{\sin A}-\frac{\cos C}{\sin C}$

$=\cot B-\cot A+\cot C-\cot B+\cot A-\cot C$

$=0$

= RHS

Hence proved.

(iv) RHS $=\sin ^{2} A+\sin ^{2}(A-B)-2 \sin A \cos B \sin (A-B)$

$=\sin ^{2} A+\sin (A-B)\{\sin (A-B)-2 \sin A \cos B\}$

$=\sin ^{2} A+\sin (A-B)(\sin A \cos B-\cos A \sin B-2 \sin A \cos B)$

$=\sin ^{2} A+\sin (A-B)(-\sin A \cos B-\cos A \sin B)$

$=\sin ^{2} A-\sin (A-B)(\sin A \cos B+\cos A \sin B)$

$=\sin ^{2} A-\sin (A-B) \sin (A+B)$

$=\sin ^{2} A-\left(\sin ^{2} A-\sin ^{2} B\right)$

$=\sin ^{2} A-\sin ^{2} A+\sin ^{2} B$

$=\sin ^{2} B$

= RHS

Hence proved.

(v) LHS $=\cos ^{2} A+\cos ^{2} B-2 \cos A \cos B \cos (A+B)$

$=\cos ^{2} A+1-\sin ^{2} B-2 \cos A \cos B \cos (A+B)$

$=1+\cos ^{2} A-\sin ^{2} B-2 \cos A \cos B \cos (A+B)$

$=1+\cos ^{2} A-\sin ^{2} B-2 \cos A \cos B \cos (A+B)$

$=1+\cos (A+B) \cos (A-B)-2 \cos A \cos B \cos (A+B)$

$=1+\cos (A+B)\{\cos (A-B)-2 \cos A \cos B\}$

$=1+\cos (A+B)(\cos A \cos B+\sin A \sin B-2 \cos A \cos B)$

$=1+\cos (A+B)(-\cos A \cos B+\sin A \sin B)$

$=1-\cos (A+B)(\cos A \cos B-\sin A \sin B)$

$=1-\cos (A+B) \cos (A+B)$

$=1-\cos ^{2}(A+B)$

$=\sin ^{2}(A+B)$

= RHS

Hence proved.

(vi) $\mathrm{LHS}=\frac{\tan (A+B)}{\cot (A-B)}$

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

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

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

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

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

$=\frac{\tan ^{2} A-\tan ^{2} B}{1-\tan ^{2} A \tan ^{2} B}$

= RHS

Hence proved.