Question

if x secϴ + y tanϴ = 1 and x tanϴ + y secϴ = m, then prove that x2-y2=1-m2

Answer 1

Given,

$x\sec\theta+y\tan\theta=1\\\\\text{Making square of both sides}\\\\(x\sec\theta+y\tan\theta)^{2}=1^{2}\\ \\x^{2}\sec^{2}\theta+y^{2}\tan^{2}\theta+2x\sec\theta.y\tan\theta=1\\\\x^{2}\sec^{2}\theta+y^{2}\tan^{2}\theta+2xy\sec\theta\tan\theta=1..........i)\\\\and,\\\\x\tan\theta+y\sec\theta=m\\\\\text{Making square of both sides}\\\\(x\tan\theta+y\sec\theta)^{2}=m^{2}\\\\x^{2}\tan^{2}\theta+y^{2}\sec\theta+2x\tan\theta+y\sec\theta=m^{2}\\\\x^{2}\tan^{2}\theta+y^{2}\sec\theta+2xy\tan\theta\sec\theta=m^{2}......ii)$

Subtracting eq ii) from i)

$(x^{2}\sec^{2}\theta+y^{2}\tan^{2}\theta+2xy\sec\theta\tan\theta)-(x^{2}\tan^{2}\theta+y^{2}\sec\theta+2xy\tan\theta\sec\theta)=1-m^{2}\\\\x^{2}(\sec^{2}\theta-\tan^{2}\theta)+y^{2}(\tan^{2}\theta-\sec^{2}\theta)=1-m^{2}\\\\x^{2}(\sec^{2}\theta-\tan^{2}\theta)-y^{2}(\sec^{2}\theta-\tan^{2}\theta)=1-m^{2}\\\\x^{2}-y^{2}=1-m^{2}\\\\\\\textbf{Note:-}\;\;\sec^{2}\theta-\tan^{2}\theta=1$