please solve
l tan θ + m sec θ = n .....(1) l' tan θ - m' sec θ = n'........(2)
mltiply 1 with m' and 2 with m and add we get sec θ term cancelled out and tan θ = (nm' + mn') / (lm' + ml')
similaly multiply 1 with l' and 2 with l and subtract tan θ cancelled out and we have sec θ = (nl' - ln') / (ml' + lm')
we knoe 1 + tan 2 θ = sec 2 θ so put the value get the answer
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