Math, asked by pri5yanjanHasminder, 1 year ago

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

Answers

Answered by MaheswariS
15

\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}}

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