Question

If l tan A + m sec A = n and l’tan A – m’sec A = n’, then show that (nl’ – ln’/ml’ + lm’)^2 = 1 + (nm’ – mn’/ lm’ + ml’) ^2

Answer 1

$\textbf{Given:}$

$l\,tanA+m\,secA=n$

$l'\,tanA-m'\,secA=n'$

$\textbf{To prove:}$

$\dfrac{(nl'-ln')^2}{(lm'+ml')^2}=1+\dfrac{(mn'+nm')^2}{(lm'+ml')^2}$

$\textbf{Solution:}$

$\text{The given equations can be written as}$

$l\,tanA+m\,secA-n=0$

$l'\,tanA-m'\,secA-n'=0$

$\text{By cross multiplication rule, we get}$

$\dfrac{tanA}{-mn'-nm'}=\dfrac{secA}{-nl'+ln'}=\dfrac{1}{-lm'-ml'}$

$\dfrac{tanA}{-(mn'+nm')}=\dfrac{secA}{-(nl'-ln')}=\dfrac{1}{-(lm'+ml')}$

$\implies$

$\dfrac{tanA}{-(mn'+nm')}=\dfrac{1}{-(lm'+ml')}$

$tanA=\dfrac{-(mn'+nm')}{-(lm'+ml')}$

$\bf\,tanA=\dfrac{mn'+nm'}{lm'+ml'}$

$\text{and}$

$\dfrac{secA}{-(nl'-ln')}=\dfrac{1}{-(lm'+ml')}$

$secA=\dfrac{-(nl'-ln')}{-(lm'+ml')}$

$\bf\,secA=\dfrac{nl'-ln'}{lm'+ml'}$

$\text{We know that,}$

$\bf\,sec^2A-tan^2A=1$

$(\dfrac{nl'-ln'}{lm'+ml'})^2-(\dfrac{mn'+nm'}{lm'+ml'})^2=1$

$\dfrac{(nl'-ln')^2}{(lm'+ml')^2}-\dfrac{(mn'+nm')^2}{(lm'+ml')^2}=1$

$\implies\boxed{\bf\dfrac{(nl'-ln')^2}{(lm'+ml')^2}=1+\dfrac{(mn'+nm')^2}{(lm'+ml')^2}}$