Question

If xsin²¢ + ycos²¢ = sin¢×cos¢ and xsin¢=ycos¢ prove that x²+y²=1​

Answer 1

CORRECT Question :–

▪︎ If $\: \: { \bold{x \: {sin}^{2} \theta + y \: {cos}^{2} \theta = sin( \theta) . cos( \theta) }} \: \:$ and $\: \: { \bold{x \: sin( \theta) = y cos( \theta) }} \: \:$ then prove that $\: \: { \bold{x + y = 1 }} \: \:$

ANSWER :–

GIVEN –

$\: \: { \huge{.}} \: \: { \bold{x \: {sin}^{2} \theta + y \: {cos}^{2} \theta = sin( \theta) . cos( \theta) }} \: \:$

$\: \: { \huge{.}} \: \: { \bold{x \: sin( \theta) = y cos( \theta) }} \: \:$

TO PROVE :–

$\: \: { \huge{.}} \: \: { \bold{{ \bold{x + y = 1 }} }} \: \:$

SOLUTION :–

• According to the given condition –

$\\ \: \implies \: { \bold{x \: {sin}^{2} \theta + y \: {cos}^{2} \theta = sin( \theta) . cos( \theta) }} \: \:$

• We should write this as –

$\\ \: \implies \: { \bold{ \dfrac{x \: {sin}^{2} \theta + y \: {cos}^{2} \theta }{ sin( \theta) .cos( \theta) }= 1}} \: \:$

$\\ \: \implies \: { \bold{ \dfrac{x \: {sin}^{2} \theta }{ sin( \theta) . cos( \theta) }{+ \dfrac{y \: {cos}^{2} \theta }{ sin( \theta) . cos( \theta) }}= 1}} \: \:$

$\\ \: \implies \: { \bold{ \dfrac{x \: {sin} \theta }{ cos( \theta) }{+ \dfrac{y \: {cos} \theta }{ sin( \theta) }} = 1}} \: \:$

• using identity –

$\\ \: \longrightarrow \: { \pink{ \bold{ \dfrac{ \: {sin} \theta }{ cos\theta} = tan( \theta) \: \: \: \: \: and \: \: \: \: \: { \dfrac{ {cos} \theta }{{ sin} \theta \: } = cot( \theta) }}}} \: \:$

• So that –

$\\ \: \implies \: { \bold{ x \: tan( \theta) + ycot( \theta) = 1 \: \: \: \: \: \: \: - - - - eq.(1)}} \: \:$

• Now using another condition –

$\\ \: \implies { \bold{x \: sin( \theta) = y cos( \theta) }} \: \:$

• So that –

$\\ \: \implies { \bold{ \dfrac{ sin \theta }{ cos{ \theta}} = \dfrac{y}{x} \: \: \: \: and \: \: \: \: \dfrac{cos \theta}{sin \theta} = \dfrac{x}{y} }} \: \:$

$\\ \: \implies { \bold{ tan{ \theta} = \dfrac{y}{x} \: \: \: \: and \: \: \: \: cot{ \theta} = \dfrac{x}{y} }} \: \:$

• Now put the values in eq.(1) –

$\\ \: \implies \: { \bold{ \cancel x \: ( \dfrac{y}{ \cancel x} )+ \cancel y( \dfrac{x}{ \cancel y}) = 1}} \: \:$

$\\ \: \implies \large { \red{ \boxed{ \bold{ x + y = 1}}}} \: \:$

$\\ \: \: \: \: \: \: \large { \boxed { \boxed{ \bold{Hence \: \: proved}}}} \: \:$