Question

If
$x = a \sinΘ \: and \: y = b \: \tanΘ \:$
,then prove that
$\frac{ {a}^{2} } {{x}^{2}} - \frac{{b}^{2} }{ {y}^{2} } = 1$
? ​

Answer 1

Answer:

$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{Qu{\blue{EstiON}}}}}}}} ⟶$

If x=a sin Θ and y=b tan Θ, then prove that

$\frac{ {a}^{2} }{ {x}^{2} } - \frac{ {b}^{2} }{ {y}^{2} } = 1$

$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{SOLU{\blue{TION}}}}}}}}⇒$

LHS=

$⇒\frac{ {a}^{2} }{ {x}^{2} } - \frac{ {b}^{2} }{{y}^{2} }$

Step-by-step explanation:

$⇒LHS \: \frac{ {a}^{2} }{ {a}^{2} \ { \sin }^{2} Θ } - \frac{ {b}^{2} }{ {b}^{2} { \tan }^{2}Θ}$

$⇒LHS \: = \frac{1}{ { \sin}^{2} Θ} - \frac{1}{ { \tan }^{2} Θ }$

$⇒LHS \: = \frac{1}{ { \sin }^{2} Θ } - \frac{1}{ { \tan }^{2} Θ }$

$⇒LHS \: = { \csc }^{2} Θ - { \cot}^{2}Θ$

$⇒LHS \: = 1 = RHS$

$\huge\underline{ \underline{ \mathbb{ { \blue{ тн{ \pink{คห{ \purple{кร \: }}}}} }}}}$

Answer 2

$\large{\underline{\bf{\green{Given:-}}}}$

✰ $x = a \sinΘ \: and \: y = b \: \tanΘ \:$

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to prove that

$\frac{ {a}^{2} } {{x}^{2}} - \frac{{b}^{2} }{ {y}^{2} } = 1$

$\huge{\underline{\bf{\red{Solution:-}}}}$

LHS:-

$: \implies \sf \frac{ {a}^{2} }{ {x}^{2} } - \frac{ {b}^{2} }{ {y}^{2} } \\ \\ \sf \: we \: know \: that \: x = a \: sin \theta \: and \: y = b\: tan \theta \\ \\ : \implies \sf \frac{ {a}^{2} }{ {a}^{2} {sin}^{2} \theta } - \frac{ {b}^{2} }{ {b}^{2} {tan}^{2} \theta } \\ \\ : \implies \sf\frac{{ \cancel{ {a}^{2} }}}{ { \cancel{{a}^{2} }}{sin}^{2} \theta } - \frac{{ \cancel{ {b}^{2}}} }{{ \cancel{ {b}^{2} }}{tan}^{2} \theta } \\ \\ : \implies \sf { \green{\frac{1}{ {sin}^{2} \theta} = cosec {}^{2} \theta }}\: \\ \\ : \implies \sf \: and \: \: { \green{\frac{1}{ {tan}^{2} \theta} = cot {}^{2} \theta }}\\ \\ : \implies \sf {cosec}^{2} \theta - {cot}^{2} \theta \\ \\ : \implies \sf{ \purple{ {cosec}^{2} \theta - cot {}^{2} \theta = 1}} </p><p>$

LHS = RHS

Hence proved.

━━━━━━━━━━━━━━━━━━━━━━━━━