Question

$ifa \sec\alpha + b \tan \alpha + c = 0 \: and \: p \sec \alpha + q \tan \alpha + r = 0.then \:prove \: that \: (br - qc {)}^{2} - (pc - ar) {}^{2} = (aq - bp) {}^{2} .$
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Answer 1

$\huge\bold{\color{green}\mid{\underline{ \underline\mathfrak\red{Answer}}}\mid}$

$\bold { \underline{ \underline{given :- }}}$

$a \sec\theta + \tan \theta + c = 0 \\ and \\ p \sec \theta + q \tan \theta + r = 0$

$\bold{ \underline{ \underline{to \: prove:-}}}$

${(br - qc)}^{2} - {(pc - ar)}^{2} = {(aq - bp)}^{2}$

$\bold { \underline{ \underline{proof :}}}$

we have,

$a \sec\theta + \tan \theta + c = 0 \\ and \\ p \sec \theta + q \tan \theta + r = 0$

solveing these two equations by the cross multiplication for sec theta and tan theta, we get

$\bold{\frac{ \sec \theta}{br - qc} = \frac{ \tan \theta}{cp - ar} = \frac{1}{aq - bp} }</p><p>$

$\bold{ \implies \sec \theta = \frac{br - cq}{aq - bp} } \\ \\ \bold{and} \\ \\ \bold{ \tan \theta = \frac{cp - ar}{aq - bp} }$

$\bold{so \: that.. \: \: \: {\sec}^{2} \theta - { \tan}^{2} \theta = 1}$

$\bold{\implies { (\frac{br - cq}{aq - bp}) }^{2} - { (\frac{cp - ar}{aq - bp} )}^{2} = 1}$

$\bold{\implies {(br - cq)}^{2} - {(cp - ar)}^{2}} = \\ \bold{ {(aq - bp)}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bold{hence, \: proved..}$