Question

(i) tan 2a - tan a = tan a sec 2a​

Answer 1

Answer:

We have to prove that :

tan2A - tanA = tanAsec2A

On taking LHS :

tan2A–tanA

=> [2tanA/1-tan²A]-tanA

=> (2tanA-tanA+tan³A)/1-tan²A

=> (tanA+tan³A)/1-tan²A

=> tanA(1+tan²A)/1-tan²A

Now, on taking RHS :

tanAsec2A

=> tanA/cos2A

As we know that :

cos2A = 1-tan²A/1+tan²A

=> tanA/(1-tan²A/1+tan²A)

=> tanA(1+tan²A)/1-tan²A

So, LHS = RHS