# Prove that:

Question:

Prove that:

(i) $\frac{\sin A+\sin 3 A+\sin 5 A}{\cos A+\cos 3 A+\cos 5 A}=\tan 3 A$

(ii) $\frac{\cos 3 A+2 \cos 5 A+\cos 7 A}{\cos A+2 \cos 3 A+\cos 5 A}=\frac{\cos 5 A}{\cos 3 A}$

(iii) $\frac{\cos 4 A+\cos 3 A+\cos 2 A}{\sin 4 A+\sin 3 A+\sin 2 A}=\cot 3 A$

(iv) $\frac{\sin 3 A+\sin 5 A+\sin 7 A+\sin 9 A}{\cos 3 A+\cos 5 A+\cos 7 A+\cos 9 A}=\tan 6 A$

(v) $\frac{\sin 5 A-\sin 7 A+\sin 8 A-\sin 4 A}{\cos 4 A+\cos 7 A-\cos 5 A-\cos 8 A}=\cot 6 A$

(vi) $\frac{\sin 5 A \cos 2 A-\sin 6 A \cos A}{\sin A \sin 2 A-\cos 2 A \cos 3 A}=\tan A$

(vii) $\frac{\sin 11 A \sin A+\sin 7 A \sin 3 A}{\cos 11 A \sin A+\cos 7 A \sin 3 A}=\tan 8 A$

(viii) $\frac{\sin 3 A \cos 4 A-\sin A \cos 2 A}{\sin 4 A \sin A+\cos 6 A \cos A}=\tan 2 A$

(ix) $\frac{\sin A \sin 2 A+\sin 3 A \sin 6 A}{\sin A \cos 2 A+\sin 3 A \cos 6 A}=\tan 5 A$

(x) $\frac{\sin A+2 \sin 3 A+\sin 5 A}{\sin 3 A+2 \sin 5 A+\sin 7 A}=\frac{\sin 3 A}{\sin 5 A}$

(xi) $\frac{\sin (\theta+\phi)-2 \sin \theta+\sin (\theta-\phi)}{\cos (\theta+\phi)-2 \cos \theta+\cos (\theta-\phi)}=\tan \theta$

Solution:

(i) Consider LHS :

$\frac{\sin A+\sin 3 A+\sin 5 A}{\cos A+\cos 3 A+\cos 5 A}$

$=\frac{\sin A+\sin 5 A+\sin 3 A}{\cos A+\cos 5 A+\cos 3 A}$

$=\frac{2 \sin \left(\frac{A+5 A}{2}\right) \cos \left(\frac{A-5 A}{2}\right)+\sin 3 A}{2 \cos \left(\frac{A+5 A}{2}\right) \cos \left(\frac{A-5 A}{2}\right)+\cos 3 A}$

$=\frac{2 \sin 3 A \cos (-2 A)+\sin 3 A}{2 \cos 3 A \cos (-2 A)+\cos 3 A}$

$=\frac{2 \sin 3 A \cos 2 A+\sin 3 A}{2 \cos 3 A \cos 2 A+\cos 3 A}$

$=\frac{\sin 3 A[2 \cos 2 A+1]}{\cos 3 A[2 \cos 2 A+1]}$

$=\tan 3 A$

= RHS

Hence, RHS = LHS.

(ii) Consider LHS :

$\frac{\cos 3 A+2 \cos 5 A+\cos 7 A}{\cos A+2 \cos 3 A+\cos 5 A}$

$=\frac{\cos 3 A+\cos 7 A+2 \cos 5 A}{\cos A+\cos 5 A+2 \cos 3 A}$

$=\frac{2 \cos \left(\frac{3 A+7 A}{2}\right) \cos \left(\frac{3 A-7 A}{2}\right)+2 \cos 5 A}{2 \cos \left(\frac{A+5 A}{2}\right) \cos \left(\frac{A-5 A}{2}\right)+2 \cos 3 A}$

$=\frac{2 \cos 5 A \cos (-2 A)+2 \cos 5 A}{2 \cos 3 A \cos (-2 A)+2 \cos 3 A}$

$=\frac{2 \cos 5 A \cos 2 A+2 \cos 5 A}{2 \cos 3 A \cos 2 A+2 \cos 3 A}$

$=\frac{2 \cos 5 A[\cos 2 A+1]}{2 \cos 3 A[\cos 2 A+1]}$

$=\frac{\cos 5 A}{\cos 3 A}$

= RHS

Hence, RHS = LHS.

(iii) Consider LHS :

$\frac{\cos 4 A+\cos 3 A+\cos 2 A}{\sin 4 A+\sin 3 A+\sin 2 A}$

$=\frac{\cos 4 A+\cos 2 A+\cos 3 A}{\sin 4 A+\sin 2 A+\sin 3 A}$

$=\frac{2 \cos \left(\frac{4 A+2 A}{2}\right) \cos \left(\frac{4 A-2 A}{2}\right)+\cos 3 A}{2 \sin \left(\frac{4 A+2 A}{2}\right) \cos \left(\frac{4 A-2 A}{2}\right)+\sin 3 A}$

$=\frac{2 \cos 3 A \cos A+\cos 3 A}{2 \sin 3 A \cos A+\sin 3 A}$

$=\frac{\cos 3 A[2 \cos A+1]}{\sin 3 A[2 \cos A+1]}$

$=\cot 3 A$

= RHS

Hence, RHS = LHS.

(iv) Consider LHS :

$\frac{\sin 3 A+\sin 5 A+\sin 7 A+\sin 9 A}{\cos 3 A+\cos 5 A+\cos 7 A+\cos 9 A}$

$=\frac{\sin 3 A+\sin 9 A+\sin 5 A+\sin 7 A}{\cos 3 A+\cos 9 A+\cos 5 A+\sin 7 A}$

$=\frac{2 \sin \left(\frac{3 A+9 A}{2}\right) \cos \left(\frac{3 A-9 A}{2}\right)+2 \sin \left(\frac{5 A+7 A}{2}\right) \cos \left(\frac{5 A-7 A}{2}\right)}{2 \cos \left(\frac{3 A+9 A}{2}\right) \cos \left(\frac{3 A-9 A}{2}\right)+2 \cos \left(\frac{5 A+7 A}{2}\right) \cos \left(\frac{5 A-7 A}{2}\right)}$

$=\frac{2 \sin 6 A \cos (-3 A)+2 \sin 6 A \cos (-A)}{2 \cos 6 A \cos (-3 A)+2 \cos 6 A \cos (-A)}$

$=\frac{2 \sin 6 A \cos 3 A+2 \sin 6 A \cos A}{2 \cos 6 A \cos 3 A+2 \cos 6 A \cos A}$

$=\frac{2 \sin 6 A[\cos 3 A+\cos A]}{2 \cos 6 A[\cos 3 A+\cos A]}$

$=\tan 6 A$

= RHS

Hence, RHS = LHS.

(v) Consider LHS :

$\frac{\sin 5 A-\sin 7 A+\sin 8 A-\sin 4 A}{\cos 4 A+\cos 7 A-\cos 5 A-\cos 8 A}$

$=\frac{\sin 5 A-\sin 7 A+\sin 8 A-\sin 4 A}{\cos 4 A-\cos 8 A+\cos 7 A-\sin 5 A}$

$=\frac{2 \sin \left(\frac{5 A-7 A}{2}\right) \cos \left(\frac{5 A+7 A}{2}\right)+2 \sin \left(\frac{8 A-4 A}{2}\right) \cos \left(\frac{8 A+4 A}{2}\right)}{-2 \sin \left(\frac{4 A+8 A}{2}\right) \sin \left(\frac{4 A-8 A}{2}\right)-2 \sin \left(\frac{7 A+5 A}{2}\right) \sin \left(\frac{7 A-5 A}{2}\right)}$

$=\frac{2 \sin (-A) \cos 6 A+2 \sin 2 A \cos 6 A}{-2 \sin 6 A \sin (-2 A)-2 \sin 6 A \sin A}$

$=\frac{-2 \sin A \cos 6 A+2 \sin 2 A c \operatorname{os} 6 A}{2 \sin 6 A \sin 2 A-2 \sin 6 A \sin A}$

$=\frac{2 \cos 6 A[\sin 2 A-\sin A]}{2 \sin 6 A[\sin 2 A-\sin A]}$

$=\cot 6 A$

= RHS

Hence, RHS = LHS.

(vi) Consider LHS :

$\frac{\sin 5 A \cos 2 A-\sin 6 A \cos A}{\sin A \sin 2 A-\cos 2 A \cos 3 A}$

Multiplying numerator and denominator by 2, we get

$=\frac{2 \sin 5 A \cos 2 A-2 \sin 6 \mathrm{~A} \cos A}{2 \sin A \sin 2 \mathrm{~A}-2 \cos 2 A \cos 3 \mathrm{~A}}$

$=\frac{\sin (5 \mathrm{~A}+2 \mathrm{~A})+\sin (5 \mathrm{~A}-2 \mathrm{~A})-\sin (6 A+\mathrm{A})-\sin (6 \mathrm{~A}-\mathrm{A})}{\cos (\mathrm{A}-2 \mathrm{~A})+\cos (\mathrm{A}+2 \mathrm{~A})-\cos (2 \mathrm{~A}+3 \mathrm{~A})-\cos (2 \mathrm{~A}-3 \mathrm{~A})}$

$=\frac{\sin 7 A+\sin 3 A-\sin 7 A-\sin 5 A}{\cos (-A)+\cos 3 A-\cos 5 A-\cos (-A)}$

$=\frac{\sin 7 A+\sin 3 A-\sin 7 A-\sin 5 A}{\cos A+\cos 3 A-\cos 5 A-\cos A}$

$=\frac{\sin 3 A-\sin 5 A}{\cos 3 A-\cos 5 A}$

$=\frac{2 \sin \left(\frac{3 A-5 A}{2}\right) \cos \left(\frac{3 A+5 A}{2}\right)}{-2 \cos \left(\frac{3 A+5 A}{2}\right) \cos \left(\frac{3 A-5 A}{2}\right)}$

$=\frac{\sin (-A) \cos 4 A}{-\cos 4 A \cos (-A)}$

$=\frac{-\sin A \cos 4 A}{-\cos 4 A \cos A}$

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

$=\tan A$

= RHS

Hence, RHS = LHS.

(vii) Consider LHS :

$\frac{\sin 11 A \sin A+\sin 7 A \sin 3 A}{\cos 2 A \sin A+\cos 6 A \sin 3 A}$

Multiplying numerator and denominator by 2, we get

$=\frac{2 \sin 11 \mathrm{~A} \sin A+2 \sin 7 A \sin 3 A}{2 \cos 11 \mathrm{~A} \operatorname{si} n \mathrm{~A}+2 \cos 7 \mathrm{~A} \sin 3 \mathrm{~A}}$

$=\frac{\cos (11 \mathrm{~A}-\mathrm{A})-\cos (11 \mathrm{~A}+\mathrm{A})+\cos (7 \mathrm{~A}-3 \mathrm{~A})-\cos (7 \mathrm{~A}+3 \mathrm{~A})}{\sin (11 \mathrm{~A}+\mathrm{A})-\sin (11 \mathrm{~A}-\mathrm{A})+\sin (7 \mathrm{~A}+3 \mathrm{~A})-\sin (7 \mathrm{~A}-3 \mathrm{~A})}$

$=\frac{\cos 10 \mathrm{~A}-\cos 12 \mathrm{~A}+\cos 4 \mathrm{~A}-\cos 10 \mathrm{~A}}{\sin 12 \mathrm{~A}-\sin 10 \mathrm{~A}+\sin 10 \mathrm{~A}-\sin 4 \mathrm{~A}}$

$=\frac{\cos 4 A-\cos 12 A}{\sin 12 A-\sin 4 A}$

$=\frac{-2 \sin \left(\frac{4 A+12 A}{2}\right) \sin \left(\frac{4 A-12 A}{2}\right)}{2 \sin \left(\frac{12 A-4 A}{2}\right) \cos \left(\frac{12 A+4 A}{2}\right)}$

$=\frac{-\sin 8 A \sin (-4 A)}{\sin 4 A \cos 8 A}$

$=\frac{\sin 8 A \sin 4 A}{\sin 4 A \cos 8 A}$

$=\tan 8 A$

= RHS

Hence, RHS = LHS.

(viii) Consider LHS :

$\frac{\sin 3 A \cos 4 A-\sin A \cos 2 A}{\sin 4 A s \text { in } A+\cos 6 A \cos A}$

Multiplying numerator and denominator by 2, we get

$=\frac{2 \sin 3 \mathrm{~A} \cos 4 \mathrm{~A}-2 \sin A \cos 2 \mathrm{~A}}{2 \sin 4 \mathrm{~A} \sin \mathrm{A}+2 \cos 6 \mathrm{~A} \cos \mathrm{A}}$

$=\frac{\sin (3 \mathrm{~A}+4 \mathrm{~A})+\sin (3 \mathrm{~A}-4 \mathrm{~A})-\sin (\mathrm{A}+2 \mathrm{~A})-\sin (\mathrm{A}-2 \mathrm{~A})}{\cos (4 \mathrm{~A}-\mathrm{A})-\cos (4 \mathrm{~A}+\mathrm{A})+\cos (6 \mathrm{~A}+\mathrm{A})+\cos (6 \mathrm{~A}-\mathrm{A})}$

$=\frac{\sin 7 A+\sin (-\mathrm{A})-\sin 3 \mathrm{~A}-\sin (-\mathrm{A})}{\cos 3 \mathrm{~A}-\cos 5 \mathrm{~A}+\cos 7 \mathrm{~A}+\cos 5 \mathrm{~A}}$

$=\frac{\sin 7 A-\sin A-\sin 3 \mathrm{~A}+\sin A}{\cos 3 \mathrm{~A}-\cos 5 \mathrm{~A}+\cos 7 \mathrm{~A}+\cos 5 \mathrm{~A}}$

$=\frac{\sin 7 A-\sin 3 A}{\cos 3 A+\cos 7 A}$

$=\frac{2 \sin \left(\frac{7 A-3 A}{2}\right) \cos \left(\frac{7 A+3 A}{2}\right)}{2 \cos \left(\frac{3 A+7 A}{2}\right) \cos \left(\frac{3 A-7 A}{2}\right)}$

$=\frac{\sin 2 A \cos 5 A}{\cos 5 A \cos (-2 A)}$

$=\frac{\sin 2 A \cos 5 A}{\cos 5 A \cos 2 A}$

$=\tan 2 A$

= RHS

Hence, RHS = LHS.

(ix) Consider LHS :

$\frac{\sin A \sin 2 A+\sin 3 A \sin 6 A}{\sin A \cos 2 A+\sin 3 A \cos 6 A}$

Multiplying numerator and denominator by 2, we get

$=\frac{2 \sin A \sin 2 A+2 \sin 3 A \sin 6 A}{2 \sin A \cos 2 A+2 \sin 3 A \cos 6 A}$

$=\frac{\cos (A-2 A)-\cos (A+2 A)+\cos (3 A-6 A)-\cos (3 A+6 A)}{\sin (A+2 A)+\sin (A-2 A)+\sin (3 A+6 A)+\sin (3 A-6 A)}$

$=\frac{\cos (-\mathrm{A})-\cos 3 \mathrm{~A}+\cos (-3 \mathrm{~A})-\cos 9 \mathrm{~A}}{\sin 3 \mathrm{~A} \sin (-A)+\sin 9 \mathrm{~A}+\sin (-3 \mathrm{~A})}$

$=\frac{\cos A-\cos 3 \mathrm{~A}+\cos 3 \mathrm{~A}-\cos 9 \mathrm{~A}}{\sin 3 \mathrm{~A}-\sin A+\sin 9 \mathrm{~A}-\sin 3 \mathrm{~A}}$

$=\frac{\cos A-\cos 9 A}{\sin 9 A-\sin A}$

$=\frac{-2 \sin \left(\frac{A+9 \mathrm{~A}}{2}\right) \sin \left(\frac{A-9 A}{2}\right)}{2 \cos \left(\frac{A+9 A}{2}\right) \sin \left(\frac{9 A-A}{2}\right)}$

$=\frac{\sin 5 A \cos 4 A}{\sin 5 A \cos (-4 A)}$

$=\tan 5 A$

= RHS

Hence, RHS = LHS.

(x) Consider LHS :

$\frac{\sin A+2 \sin 3 A+\sin 5 A}{\sin 3 A+2 \sin 5 A+\sin 7 A}$

$=\frac{\sin A+\sin 5 A+2 \sin 3 A}{\sin 3 A+\sin 7 A+2 \sin 5 A}$

$=\frac{2 \sin \left(\frac{A+5 A}{2}\right) \cos \left(\frac{A-5 A}{2}\right)+2 \sin 3 A}{2 \sin \left(\frac{3 A+7 A}{2}\right) \cos \left(\frac{3 A-7 A}{2}\right)+2 \sin 5 A}$

$=\frac{2 \sin 3 A \cos (-2 A)+2 \sin 3 A}{2 \sin 5 A \cos (-2 A)+2 \sin 5 A}$

$=\frac{2 \sin 3 A \cos 2 A+2 \sin 3 A}{2 \sin 5 A \cos 2 A+2 \sin 5 A}$

$=\frac{2 \sin 3 A[\cos 2 A+1]}{2 \sin 5 A[\cos 2 A+1]}$

$=\frac{\sin 3 A}{\sin 5 A}$

= RHS

Hence, RHS = LHS.

(xi) Consider LHS :

$\frac{\sin (\theta+\phi)-2 \sin \theta+\sin (\theta-\phi)}{\cos (\theta+\phi)-2 \cos \theta+\cos (\theta-\phi)}$

$=\frac{\sin (\theta+\phi)+\sin (\theta-\phi)-2 \sin \theta}{\cos (\theta+\phi)+\cos (\theta-\phi)-2 \cos \theta}$

$=\frac{2 \sin \left(\frac{\theta+\phi+\theta-\phi}{2}\right) \cos \left(\frac{\theta+\phi-\theta+\phi}{2}\right)-2 \sin \theta}{2 \cos \left(\frac{\theta+\phi+\theta-\phi}{2}\right) \cos \left(\frac{\theta+\phi-\theta+\phi}{2}\right)-2 \cos \theta}$

$=\frac{2 \sin \theta \cos \phi-2 \sin \theta}{2 \cos \theta \cos \phi-2 \cos \theta}$

$=\frac{2 \sin \theta[\cos \phi-1]}{2 \cos \theta[\cos \phi-1]}$

$=\tan \theta$

= RHS

Hence, RHS = LHS.