Question

Tan(a-b)+tanb/1-tan(a-b)tanb=tana

Answer 1

Proof:

R.H.S:-

tanA = tan[(A-B)+B]

=$\frac{tan(A-B)+tanB}{1-tan(A-B)tanB}$ (from the tan(C+D) identity)

=L.H.S

Answer 2

Step-by-step explanation:

tan(a-b)+tanb/1-tan(a-b)tanb

={tana-tanb/1+tanatanb}+tanb/ 1-{tana-tanb/1+tanatanb}tanb

=tana-tanb+tanb(1+tanatanb)/ 1+tanatanb-tanb(tana-tanb)

=tana-tanb+tanb+tanatan²b/ 1+tanatanb-tanatanb+tan²b

=tana(1+tan²b)/1+tan²b

=tana