[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \dfrac{xsin\phi}{a} - \dfrac{ycos\phi}{b} = 1 \\ [/tex]
On squaring both sides, we get
[tex]\sf \: \left(\dfrac{xsin\phi}{a} - \dfrac{ycos\phi}{b}\right)^{2} = 1 \\ [/tex]
[tex]\sf \: \left(\dfrac{xsin\phi}{a}\right)^{2} + {\left(\dfrac{ycos\phi}{b} \right)}^{2} - 2 \times \dfrac{xsin\phi}{a} \times \dfrac{ycos\phi}{b} = 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} {sin}^{2}\phi}{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\phi}{ {b}^{2} } - \dfrac{2xysin\phi \: cos\phi}{ab} = 1 - - - (1)[/tex]
Further given that,
[tex]\sf \: \dfrac{xcos\phi}{a} + \dfrac{ysin\phi}{b} = 1 \\ [/tex]
[tex]\sf \: \left(\dfrac{xcos\phi}{a} + \dfrac{ysin\phi}{b}\right)^{2} = 1 \\ [/tex]
[tex]\sf \: \left(\dfrac{xcos\phi}{a} \right)^{2} + {\left(\dfrac{ysin\phi}{b}\right)}^{2} + 2 \times \dfrac{xcos\phi}{a} \times \dfrac{ysin\phi}{b} = 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} {cos}^{2}\phi}{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\phi}{ {b}^{2} } + \dfrac{2xysin\phi \: cos\phi}{ab} = 1 - - - (2)[/tex]
On adding equation (1) and (2), we get
[tex]\sf \: \dfrac{ {x}^{2} }{ {a}^{2} }( {sin}^{2}\phi + {cos}^{2}\phi) + \dfrac{ {y}^{2} }{ {b}^{2} }( {cos}^{2}\phi + {sin}^{2} \phi) = 1 + 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} }{ {a}^{2} }( 1) + \dfrac{ {y}^{2} }{ {b}^{2} }( 1) = 1 + 1 \\ [/tex]
[tex]\implies\sf \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 2 \\ [/tex]
Hence,
[tex]\implies\sf \:\boxed{\bf \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 2 \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Identities Used:
[tex]\sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ [/tex]
[tex]\sf \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ [/tex]
[tex]\sf \: {sin}^{2}x + {cos}^{2}x = 1 \\ [/tex]
Additional Information
[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\[/tex]
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−
=
1
a
xsinϕ
b
ycosϕ
=1
(
)
2
+
×
+(
−2×
x
sin
ϕ
y
cos
ab
2xysinϕcosϕ
=1−−−(1)
xcosϕ
ysinϕ
+2×
=1−−−(2)
(sin
ϕ+cos
ϕ)+
(cos
ϕ+sin
ϕ)=1+1
(1)+
(1)=1+1
⟹
=2
(x+y)
=x
+y
+2xy
(x−y)
−2xy
x+cos
x=1
‾
★
90
°
MoreFormulae
★sinx=
cosecx
★cosx=
secx
★tanx=
cosx
sinx
cotx
★cotx=
tanx
★cosecx=
★secx=
★sin
★sec
x−tan
★cosec
x−cot
★sin(90°−x)=cosx
★cos(90°−x)=sinx
★tan(90°−x)=cotx
★cot(90°−x)=tanx
★cosec(90°−x)=secx
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \dfrac{xsin\phi}{a} - \dfrac{ycos\phi}{b} = 1 \\ [/tex]
On squaring both sides, we get
[tex]\sf \: \left(\dfrac{xsin\phi}{a} - \dfrac{ycos\phi}{b}\right)^{2} = 1 \\ [/tex]
[tex]\sf \: \left(\dfrac{xsin\phi}{a}\right)^{2} + {\left(\dfrac{ycos\phi}{b} \right)}^{2} - 2 \times \dfrac{xsin\phi}{a} \times \dfrac{ycos\phi}{b} = 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} {sin}^{2}\phi}{ {a}^{2} } + \dfrac{ {y}^{2} {cos}^{2}\phi}{ {b}^{2} } - \dfrac{2xysin\phi \: cos\phi}{ab} = 1 - - - (1)[/tex]
Further given that,
[tex]\sf \: \dfrac{xcos\phi}{a} + \dfrac{ysin\phi}{b} = 1 \\ [/tex]
On squaring both sides, we get
[tex]\sf \: \left(\dfrac{xcos\phi}{a} + \dfrac{ysin\phi}{b}\right)^{2} = 1 \\ [/tex]
[tex]\sf \: \left(\dfrac{xcos\phi}{a} \right)^{2} + {\left(\dfrac{ysin\phi}{b}\right)}^{2} + 2 \times \dfrac{xcos\phi}{a} \times \dfrac{ysin\phi}{b} = 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} {cos}^{2}\phi}{ {a}^{2} } + \dfrac{ {y}^{2} {sin}^{2}\phi}{ {b}^{2} } + \dfrac{2xysin\phi \: cos\phi}{ab} = 1 - - - (2)[/tex]
On adding equation (1) and (2), we get
[tex]\sf \: \dfrac{ {x}^{2} }{ {a}^{2} }( {sin}^{2}\phi + {cos}^{2}\phi) + \dfrac{ {y}^{2} }{ {b}^{2} }( {cos}^{2}\phi + {sin}^{2} \phi) = 1 + 1 \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} }{ {a}^{2} }( 1) + \dfrac{ {y}^{2} }{ {b}^{2} }( 1) = 1 + 1 \\ [/tex]
[tex]\implies\sf \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 2 \\ [/tex]
Hence,
[tex]\implies\sf \:\boxed{\bf \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 2 \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Identities Used:
[tex]\sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ [/tex]
[tex]\sf \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ [/tex]
[tex]\sf \: {sin}^{2}x + {cos}^{2}x = 1 \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\[/tex]
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ab
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On adding equation (1) and (2), we get
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Identities Used:
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MoreFormulae
MoreFormulae
★sinx=
cosecx
1
★cosx=
secx
1
★tanx=
cosx
sinx
=
cotx
1
★cotx=
sinx
cosx
=
tanx
1
★cosecx=
sinx
1
★secx=
cosx
1
★sin
2
x+cos
2
x=1
★sec
2
x−tan
2
x=1
★cosec
2
x−cot
2
x=1
★sin(90°−x)=cosx
★cos(90°−x)=sinx
★tan(90°−x)=cotx
★cot(90°−x)=tanx
★cosec(90°−x)=secx