If LsecФ - MtanФ + n = 0 and L'secФ - M'tanФ + n = 0
Then Prove that : (M'N + MN')²-(N'L - NL')² = (L'M - LM')²
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There is a mistake in the question... LHS should have: (M'N - MN')²
Let Ф = A for easy writing.
L secA - M tan A = - N L' secA - M' tan A = - N'
M' N - M N' = sec A (M' L - M L')
N' L - N L' = LM' tan A - LL' sec A + LL' sec A - ML' tanA
= tanA (LM' - ML')
LHS = (LM' - ML')² (sec²A - tan²A)
= RHS
Let Ф = A for easy writing.
L secA - M tan A = - N L' secA - M' tan A = - N'
M' N - M N' = sec A (M' L - M L')
N' L - N L' = LM' tan A - LL' sec A + LL' sec A - ML' tanA
= tanA (LM' - ML')
LHS = (LM' - ML')² (sec²A - tan²A)
= RHS
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