Question

If x = asinθ + bcosθ and y = acosθ - bsinθ , prove that x2 + y2 = a2 + b2

Answer 1

GIVEN :-

$\\ \bullet \sf \:x = asin \theta + bcos\theta \\ \\ \bullet \sf \: y = acos\theta - sin\theta$

TO PROVE :-

$\\ \bullet \sf \: {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2}$

SOLUTION :-

$\\ \sf \: L.H.S = {x}^{2} + {y}^{2}$

Putting values of x and y ,

$\\ \implies \sf \: (asin\theta + bco {s\theta)}^{2} + (acos\theta - bsi {n\theta)}^{2} \\ \\ \\ \implies \sf \: {a}^{2} {sin}^{2}\theta + {b}^{2} {cos}^{2}\theta + 2(asin\theta)(bcos\theta) + \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: {a}^{2} {cos}^{2}\theta + {b}^{2} {sin}^{2}\theta - 2(acos\theta)(bsin\theta)$

$\\ \implies \sf \: {a}^{2} {sin}^{2}\theta + {a}^{2} {cos}^{2}\theta + {b}^{2} {cos}^{2}\theta + {b}^{2} {sin}^{2}\theta + \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \cancel{2absin\theta.cos\theta} - \cancel{2absin\theta .cos\theta}$

Taking a² common and b² common respectively,

$\\ \implies \sf \: {a}^{2} ( {sin}^{2}\theta + {cos}^{2}\theta ) + {b}^{2} ( {cos}^{2}\theta + {sin}^{2}\theta ) \:$

★ sin²ø + cos²ø = 1

$\\ \sf \implies \: {a}^{2} (1) + {b}^{2} (1) \\ \\ \implies \sf \: {a}^{2} + {b}^{2} \: \: \: \: \: = R.H.S$

Hence ,

$\\ \underline{\overline{\boxed{ \sf {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2} }}}$

MORE IDENTITIES :-

♦ 1 + tan²ø = sec²ø

♦ 1 + cot²ø = cosec²ø

♦ sin2ø = 2sinø.cosø

♦ cos2ø = cos²ø - cos²ø

ㅤㅤㅤㅤ= 2cos²ø - 1

ㅤㅤㅤㅤ= 1 - 2sin²ø

Answer 2

Answer:

GIVEN :-

\begin{gathered} \\ \bullet \sf \:x = asin \theta + bcos\theta \\ \\ \bullet \sf \: y = acos\theta - sin\theta \\ \\ \end{gathered}

∙x=asinθ+bcosθ

∙y=acosθ−sinθ

TO PROVE :-

\begin{gathered} \\ \bullet \sf \: {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2} \\ \\ \end{gathered}

∙x

2

+y

2

=a

2

+b

2

SOLUTION :-

\begin{gathered} \\ \sf \: L.H.S = {x}^{2} + {y}^{2} \\ \end{gathered}

L.H.S=x

2

+y

2

Putting values of x and y ,

\begin{gathered} \\ \implies \sf \: (asin\theta + bco {s\theta)}^{2} + (acos\theta - bsi {n\theta)}^{2} \\ \\ \\ \implies \sf \: {a}^{2} {sin}^{2}\theta + {b}^{2} {cos}^{2}\theta + 2(asin\theta)(bcos\theta) + \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: {a}^{2} {cos}^{2}\theta + {b}^{2} {sin}^{2}\theta - 2(acos\theta)(bsin\theta) \\ \end{gathered}

⟹(asinθ+bcosθ)

2

+(acosθ−bsinθ)

2

⟹a

2

sin

2

θ+b

2

cos

2

θ+2(asinθ)(bcosθ)+

a

2

cos

2

θ+b

2

sin

2

θ−2(acosθ)(bsinθ)

\begin{gathered} \\ \implies \sf \: {a}^{2} {sin}^{2}\theta + {a}^{2} {cos}^{2}\theta + {b}^{2} {cos}^{2}\theta + {b}^{2} {sin}^{2}\theta + \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \cancel{2absin\theta.cos\theta} - \cancel{2absin\theta .cos\theta}\\ \end{gathered}

⟹a

2

sin

2

θ+a

2

cos

2

θ+b

2

cos

2

θ+b

2

sin

2

θ+

2absinθ.cosθ

−

2absinθ.cosθ

Taking a² common and b² common respectively,

\begin{gathered} \\ \implies \sf \: {a}^{2} ( {sin}^{2}\theta + {cos}^{2}\theta ) + {b}^{2} ( {cos}^{2}\theta + {sin}^{2}\theta ) \: \\ \\ \end{gathered}

⟹a

2

(sin

2

θ+cos

2

θ)+b

2

(cos

2

θ+sin

2

θ)

★ sin²ø + cos²ø = 1

\begin{gathered} \\ \sf \implies \: {a}^{2} (1) + {b}^{2} (1) \\ \\ \implies \sf \: {a}^{2} + {b}^{2} \: \: \: \: \: = R.H.S \\ \\ \end{gathered}

⟹a

2

(1)+b

2

(1)

⟹a

2

+b

2

=R.H.S

Hence ,

\begin{gathered} \\ \underline{\overline{\boxed{ \sf {x}^{2} + {y}^{2} = {a}^{2} + {b}^{2} }}} \\ \\ \\ \end{gathered}

x

2

+y

2

=a

2

+b

2

MORE IDENTITIES :-

♦ 1 + tan²ø = sec²ø

♦ 1 + cot²ø = cosec²ø

♦ sin2ø = 2sinø.cosø

♦ cos2ø = cos²ø - cos²ø

ㅤㅤㅤㅤ= 2cos²ø - 1

ㅤㅤㅤㅤ= 1 - 2sin²ø

Step-by-step explanation:

Hope it helps

Good night