Question

If sec 0 + tan 0 = m and sec 0 -tan 0 = n, find the value of mn.​

Answer 1

Answer:

m=sec0 + tan 0

n= sec0 - tan 0

mn= (sec 0+ tan 0)(sec0- tan 0)

(a+b)(a-b)=a^2-b^2

where,

a=sec0 + tan 0

b=sec0 - tan 0

so,

mn= (sec 0+ tan 0)(sec0- tan 0)

=sec^2 0 - tan^2 0

identity,

sec^2 0 - tan^2 0 = 1

so, sec^2 0 - tan^2 0 = 1

1 is your required answer

Step-by-step explanation:

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Answer 2

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

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

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

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

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

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