Question

Prove the following: (tan A - tan B)^2 + (1 + tan A tanB)^2= sec^2 A sec^2 B​

Answer 1

Answer:

Step-by-step explanation:

[tanA - tanB ]^2 + [1+tan A tan B]^2 ===

tan^2A - 2tanAtanB+ tan^2B     +     1 +2tanAtanB + tan^2Atan^2B  ===

then -2tanAtanB and +2tanAtanB get cancelled

thus     tan^2A       +[ tan^2B +1]      +tan^2Atan^2B =

tan^2B +1 = sec^2B

then take tan^2A common and you get

tan^2A[  [ 1+ tan^2B ]  +sec^2B

again you get

tan^2A x sec^2B + sec^2B

then take sec^2B common

thus you get        sec^2B [ 1 +tan^2A ]

then again    sec^2B x sec^2A

therefore LHS = RHS

hence proved