Question

13. Prove that: tan 74 tan 4A tan 3A = tan 74 -tan 4A -tan 3A.​

Answer 1

Answer:

Step-by-step explanation:

$tan{7A} = tan{(3A + 4A)}\\or, tan{7A} = \frac{tan{3A} + tan{4A}}{1-tan{3A}\cdot tan{4A}}\\or,tan{7A}-tan{7A}\cdot tan{3A}\cdot tan{4A} = tan{3A} + tan{4A}\\or, tan{7A} -tan{3A} - tan{4A} = tan{7A}\cdot tan{3A}\cdot tan{4A}$