Question

k) 1+tan 4A. tan 2A = sec 4A​

Answer 1

Step-by-step explanation:

to prove 1+tan 4A. tan 2A = sec 4A​

LHS = 1+tan 4A. tan 2A

       = 1 + $\frac{sin4A}{cos4A}$*$\frac{sin2A}{cos2A}$

       = $\frac{cos4Acos2A+sin4Asin2A}{cos4Acos2A}$

       = $\frac{cos(4A-2A)}{cos4A*cos2A}$

       = $\frac{cos2A}{cos4A*cos2A}$

       = $\frac{1}{cos4A}$

       = sec4A

       = RHS

=> LHS = RHS

Hence proved