Question

if sec theta + tan theta = m and sec theta - tan theta = n then find the value of mn​

Answer 1

Answer:

$mn \\ = ( \sec(x) + \tan(x) )( \sec(x) - \tan(x) ) \\ = \sec {}^{2} (x) - \tan {}^{2} (x) \\ = 1$

Answer 2

$\rm{m\times n\:is\:equal\:to\:(sec\:\theta+tan\:\theta)(sec\:\theta-tan\:\theta)}$

$\longrightarrow\rm{mn =(sec\:\theta+tan\:\theta)(sec\:\theta-tan\:\theta) }$

$\longrightarrow\rm{mn =sec^2\:\theta-tan^2\:\theta\quad...\:Since\:(a+b)(a-b)=a^2-b^2}$

$\longrightarrow\rm{mn =1/\cos ^{2} \theta-\sin ^{2} \theta/\cos ^{2} \theta\quad...\:Since\:sec\:\theta=1/\cos\theta\:\&\: tan\:\theta=\sin\:\theta/\cos\:\theta}$

$\longrightarrow\rm{mn =(1-\sin ^{2} \theta)/\cos ^{2} \theta=\cos ^{2} \theta/\cos ^{2} \theta}$

$\longrightarrow\rm{mn =1\quad...\:Hence\:Proved!}$