Question

Prove that: (tan A - tan B)² + (1 + tan A tan B)² = sec² A sec² B

Answer 1

$\huge{\pink{\underline{\ulcorner{\star\:Solution\: \star}\urcorner}}}$

Taking LHS from given equation:

$\implies tan^{2}A+tan^{2}B -\cancel{2tan A tanB } + 1 + tan^{2}A tan^{2}B + \cancel{2tanA tan B }\\\implies tan^{2}A + tan^{2}B +1 +tan^{2}Atan^{2}B\\\implies 1(tan^{2}A +1 )+tan^{2}B(1+tan^{2}A)\\\implies (tan^{2}A+1)(tan^{2}B+1)\\\implies sec^{2}A.sec^{2}B$

We have ,

$\therefore tan^{2}\theta+1= sec^{2}\theta$

Now the answer is

L.H.S = RHS

$\red{\huge{Hence\:Proved}}$