Match the following columns with the values obtained for the solution- I- xcos- ysin-a- xsin- ycos-b a x2-y22-16xy II- x- -tan- y-tan- b xy - 1 III- x- ytan-a- xtan- y-b c x2-y2-a2-b2 IV- x-cos- y-cos- d x2-y2-a2-b2

The correct option is B I d, II b, III c, IV a I. #160; xcosθ #160; +ysinθ #160; =a.....(i) xsinθ #160; ycosθ #160; =b.....(ii) S ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Match the fol...

Question

Match the following columns with the values obtained for the solution.
$I .$
$x cos θ + y sin θ = a$ ,
$x sin θ - y cos θ = b$

$a)$ $(x^{2} - y^{2})^{2} = 16 x y$
$I I .$
$x = sec θ + tan θ$ ,
$y = sec θ - tan θ$
$b)$ $x y = 1$
$I I I .$
$x sec θ + y tan θ = a$ ,
$x tan θ + y sec θ = b$

$c)$ $x^{2} - y^{2} = a^{2} - b^{2}$

$I V .$
$x = cot θ + cos θ$ ,
$y = cot θ - cos θ$

$d)$ $x^{2} + y^{2} = a^{2} + b^{2}$

I - c, I I - d, I I I - a, I V - b

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I - d, I I - b, I I I - c, I V - a

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I - a, I I - c, I I I - d, I V - b

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I - b, I I - c, I I I - a, I V - d

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Solution

The correct option is B $I - d, I I - b, I I I - c, I V - a$
I. $x cos θ + y sin θ = a . . . . . (i)$
$x sin θ - y cos θ = b . . . . . (i i)$
Squaring and adding equations $(i)$ and $(i i)$ , we get
${[x cos θ + y sin θ]}^{2} + {[x sin θ - y cos θ]}^{2} = a^{2} + b^{2}$
$x^{2} {cos}^{2} θ + y^{2} {sin}^{2} θ + 2 x y sin θ cos θ + x^{2} {sin}^{2} θ + y^{2} {cos}^{2} θ - 2 x y sin θ cos θ = a^{2} + b^{2}$
$x^{2} [{sin}^{2} θ + {cos}^{2} θ] + y^{2} [{sin}^{2} θ + {cos}^{2} θ] = a^{2} + b^{2}$
$x^{2} + y^{2} = a^{2} + b^{2}$
which is (d).
II. $x = sec θ + tan θ$
$y = sec θ - tan θ$
$x y = (sec θ + tan θ) (sec θ - tan θ)$
$x y = {sec}^{2} θ - {tan}^{2} θ$
$x y = 1 [∵ 1 + {tan}^{2} θ = {sec}^{2} θ]$
which is (b).
III. $x sec θ + y tan θ = a$ ....... $(i i i)$
$x tan θ + y sec θ = b$ ......... $(i v)$
Squaring and then subtracting equation $(i i i)$ from $(i v)$ , we get
$[x^{2} {sec}^{2} θ + y^{2} {tan}^{2} θ + 2 x y sec θ tan θ] - [x^{2} {tan}^{2} θ + y^{2} {sec}^{2} θ + 2 x y sec θ tan θ] = a^{2} - b^{2}$
$x^{2} [{sec}^{2} θ - {tan}^{2} θ] - y^{2} [{sec}^{2} θ - {tan}^{2} θ] = a^{2} - b^{2}$
$x^{2} - y^{2} = a^{2} - b^{2}$
which is (c).
IV. $x = cot θ + cos θ$
$y = cot θ - cos θ$
Squaring both equations, we get
$x^{2} = {cot}^{2} θ + {cos}^{2} θ + 2 cot θ cos θ$ ......... $(v)$
$y^{2} = {cot}^{2} θ + {cos}^{2} θ - 2 cot θ cos θ$ ......... $(v i)$
Subtract equation $(v i)$ from $(v)$ , we get
$x^{2} - y^{2} = 4 cot θ cos θ$
Squaring both sides, we get
${(x^{2} - y^{2})}^{2} = 16 {cot}^{2} θ {cos}^{2} θ$ ......... $(v i i)$
Now, $x y = (cot θ + cos θ) (cot θ - cos θ)$
$x y = {cot}^{2} θ - {cos}^{2} θ$
$x y = \frac{{cos}^{2} θ}{{sin}^{2} θ} [1 - {sin}^{2} θ]$
$x y = {cot}^{2} θ {cos}^{2} θ$ ......... $(v i i i)$
Substitute $(v i i i)$ in equation $(v i i)$ , we get
${(x^{2} - y^{2})}^{2} = 16 x y$
which is (a).
Therefore, I-d, II-b, III-c, IV-a

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$I .$ $x cos θ + y sin θ = a$ , $x sin θ - y cos θ = b$	$a)$ $(x^{2} - y^{2})^{2} = 16 x y$
$I I .$ $x = sec θ + tan θ$ , $y = sec θ - tan θ$	$b)$ $x y = 1$
$I I I .$ $x sec θ + y tan θ = a$ , $x tan θ + y sec θ = b$	$c)$ $x^{2} - y^{2} = a^{2} - b^{2}$
$I V .$ $x = cot θ + cos θ$ , $y = cot θ - cos θ$	$d)$ $x^{2} + y^{2} = a^{2} + b^{2}$

Match the following columns with the values obtained for the solution.I.xcosθ+ysinθ=a,xsinθ−ycosθ=ba) (x2−y2)2=16xy II.x=secθ+tanθ,y=secθ−tanθb) xy=1III.xsecθ+ytanθ=a,xtanθ+ysecθ=b c) x2−y2=a2−b2IV. x=cotθ+cosθ,y=cotθ−cosθ d) x2+y2=a2+b2