Question

If
$x \cos(theta) = a$
and
$y = a \tan(theta)$
, then prove that
${x}^{2} - {y}^{2} = {a}^{2}$
​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Given:-}$

$x \cos \theta = a$

$\implies x = \frac{a}{ \cos\theta }$

$\implies x = a \sec \theta$

Now,

$y = a \tan \theta$

$\star\:\:\:\bf\large\underline\blue{To\:Prove:-}$

• ${x}^{2} - {y}^{2} = {a}^{2}$

$\star\:\:\:\bf\underline\blue{LHS:-}$

${x}^{2} - {y}^{2}$

$= {(a \sec \theta)}^{2} - {(a \tan \theta)}^{2}$

$= a^{2} \sec^{2} \theta - a^{2} \tan^{2} \theta$

$= a^{2} (\sec^{2} \theta -\tan^{2} \theta)$

Now, we know that,

$(\sec^{2} \theta -\tan^{2} \theta) = 1$

So,

$a^{2} (\sec^{2} \theta -\tan^{2} \theta)$

$= {a}^{2} \times 1$

$= {a}^{2}$

$\star\:\:\:\bf\underline\blue{RHS:-}$

$= {a}^{2}$

$\star\:\:\:\bf\underline\blue{Therefore,\:LHS\:=\:RHS}$

$\star\:\:\:\bf\large\underline\blue{\textsf{Hence Proved.}}$