Prove that: (i) (sin 3A + sin 5A + sin 7A + sin 9A)/(cos 3A + cos 5A + cos 7A + cos 9A) = tan 6A

Question

Prove that:

(i) \({\frac{sin 3A + sin 5A + sin 7A + sin 9A}{cos3A + cos5A + cos7A + cos 9A}}\) = tan 6A

(ii) \({\frac{sin 5A - sin7A + sin8A - sin 4A}{cos 4A + cos7A - cos 5A - cos8A}}\) = cot 6A

(iii) \({\frac{sin 5Acos2A - sin 6A cos A}{sin A sin 2A - cos2A cos 3A}}\) = tan A

Answer 1

(i) Let us consider the LHS

\({\frac{sin 3A + sin 5A + sin 7A + sin 9A}{cos3A + cos5A + cos7A + cos 9A}}\)

On using the formulas,

cos A + cos B = 2cos(A + B)/2 cos(A - B)/2

sin A + sin B = 2 sin(A + B)/2 cos(A - B)/2

Therefore now,

= RHS

Thus proved.

(ii) Let us consider the LHS

\({\frac{sin 5A - sin7A + sin8A - sin 4A}{cos 4A + cos7A - cos 5A - cos8A}}\)

On using the formulas,

cos A - cos B = -2 sin(A + B)/2 sin(A - B)/2

sin A - sin B = 2 cos(A + B)/2 sin(A - B)/2

Therefore, now

= cot 6A

= RHS

Thus proved.

(iii) Let us the consider the LHS

On using the formulas,

= tan A

= RHS

Thus proved.