Question

If x = a sec theta + b tan theta n y = a tan theta + b sec theta then proove tat :
$x { }^{2} - y {}^{2} = a {}^{2} - b {}^{2}$

Answer 1

X=asec+btan
Y=atan+bsec
${x}^{2} - {y}^{2} = {(asec + btan)}^{2} - {(atan + bsec)}^{2} \\ {a}^{2} . {sec}^{2} + {b}^{2} . {tan}^{2} + 2.a.b.sec.tan - ( {a}^{2}. {tan}^{2} + {b}^{2}. {sec}^{2} + 2.a.b.sec.tan \\ {a}^{2} {sec}^{2} + {b}^{2}. {tan}^{2} - {b}^{2}. {sec}^{2} - {a}^{2}. {tan}^{2} \\ {a}^{2} ( {sec}^{2} - {tan}^{2} ) + {b}^{2} ( {tan}^{2} - {sec}^{2} ) \\ {a}^{2} .(1) - {b}^{2} .( {sec}^{2} - {tan}^{2} ) \\ = {a}^{2} - {b}^{2} \\ lhs = rhs \\ hence \: proved$

Answer 2

hence prove you are satisfied