Question

if cosec∅ - sec∅= l sec∅- cos∅=m , then what is the value of l^2m^2[l^2+m^2+3]

Answer 1

marknit ss brainliest

Answer 2

cosec∅ - sin∅= l
1/sin ∅ -sin ∅ =l
1-sin^2 ∅ /sin ∅ =l
Cos^2 ∅ /sin ∅ =l

sec∅- cos∅=m
1/cos ∅ - cos ∅ =m
1-cos^2 ∅/cos ∅ =m
Sin^2 ∅ /cos ∅ =m

Now,
l^2m^2[l^2+m^2+3]

(Cos^2 ∅ /sin ∅) ^2 × (Sin^2 ∅ /cos ∅) ^2 [ (Cos^2 ∅ /sin ∅) ^2 + (Sin^2 ∅ /cos ∅) ^2 +3]

=Cos^4 ∅ /sin^2 ∅ ×sin^4 ∅ /cos^2 ∅ [ Cos^4 ∅ /sin^2 ∅ +sin^4 ∅ /cos^2 ∅+3]

=cos^2 ∅sin^2 ∅[cos^6 ∅+sin^6 ∅+3 sin^2 ∅cos^2 ∅ /sin^2 ∅cos^2 ∅]

= cos^2 ∅sin^2 ∅[(cos^2 ∅) ^3+(sin^2 ∅) ^3+3 sin^2 ∅cos^2 ∅/ sin^2 ∅cos^2 ∅]

= cos^2 ∅sin^2 ∅[(cos ^2 ∅+sin^2 ∅)^3-3 sin^2 ∅cos^2 ∅ + 3 sin^2 ∅cos^2 ∅ /sin^2 ∅cos^2 ∅]

= cos^2 ∅sin^2 ∅.1/ sin^2 ∅cos^2 ∅

= cos^2 ∅sin^2 ∅/ cos^2 ∅sin^2 ∅

=1

Hope it helps