Question

(1+tan A. tan B)²+ (tan A-tan B)²= sec² A sec²B​

Answer 1

Answer:

$1 + {(tana \: tanb)}^{2} + 2tana \: tanb \: + {tan}^{2}a + {tan}^{2} b - 2tana \: tanb \\ 1 + {tan}^{2} a \: {tan}^{2} b \: + {tan}^{2} a + \: {tan}^{2} b \\ 1 + {tan}^{2} a + {tan}^{2} b + {tan}^{2} a \: {tan}^{2} b \\ 1 + {tan}^{2} a + {tan}^{2} b(1 + {tan}^{2} a) \\ (1 + {tan}^{2} a)(1 + {tan}^{2} b) \\ ({sec}^{2} a)( {sec}^{2} b)$

Hence LHS=RHS

hope it helps...............