Question

a sec theta + b tan theta + c= 0 and p sec theta + q tan theta + r= 0. prove that (br-qc)^2 - ( pc- at)^2=(aq-bq)^2
​

Answer 1

Answer:

Refer to the above attachment bro.......

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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