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Question Number 57383 by Tinkutara last updated on 03/Apr/19

Commented by Tinkutara last updated on 03/Apr/19
B part
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19
	$cos^(−1) (y)+cos^(−1) (bxy)=(π/2)−sin^(−1) (ax) cos{cos^(−1) (y)+cos^(−1) (bxy)}=cos((π/2)−sin^(−1) ax) y×bxy−(√(1−y^2 )) ×(√(1−b^2 x^2 y^2 )) =ax bxy^2 −ax=(√(1−y^2 )) ×(√(1−b^2 x^2 y^2 )) b^2 x^2 y^4 +a^2 x^2 −2abx^2 y^2 =1−b^2 x^2 y^2 −y^2 +b^2 x^2 y^4 a^2 x^2 −2abx^2 y^2 −1+b^2 x^2 y^2 +y^2 =0 now a=1 b=0 x^2 +y^2 =1→circle when a=1 b=1 x^2 −2x^2 y^2 −1+x^2 y^2 +y^2 =0 x^2 −x^2 y^2 −1+y^2 =0 x^2 (1−y^2 )−(1−y^2 )=0 (1−y^2 )(x^2 −1)=0 (x^2 −1)(y^2 −1)=0 a=1 b=2 x^2 −4x^2 y^2 −1+4x^2 y^2 +y^2 =0 x^2 +y^2 =1 a=2 b=2 4x^2 −8x^2 y^2 −1+4x^2 y^2 +y^2 =0 4x^2 −4x^2 y^2 −1+y^2 =0 4x^2 (1−y^2 )−(1−y^2 )=0 (1−y^2 )(4x^2 −1)=0 (y^2 −1)(4x^2 −1)=0 plscheck..$
	${c o s}^{- 1} (y) + {c o s}^{- 1} (b x y) = \frac{π}{2} - {s i n}^{- 1} (a x)$ $c o s {{c o s}^{- 1} (y) + {c o s}^{- 1} (b x y)} = c o s (\frac{π}{2} - {s i n}^{- 1} a x)$ $y \times b x y - \sqrt{1 - y^{2}} \times \sqrt{1 - b^{2} x^{2} y^{2}} = a x$ ${b x y}^{2} - a x = \sqrt{1 - y^{2}} \times \sqrt{1 - b^{2} x^{2} y^{2}}$ $b^{2} x^{2} y^{4} + a^{2} x^{2} - 2 {a b x}^{2} y^{2} = 1 - b^{2} x^{2} y^{2} - y^{2} + b^{2} x^{2} y^{4}$ $a^{2} x^{2} - 2 {a b x}^{2} y^{2} - 1 + b^{2} x^{2} y^{2} + y^{2} = 0$ $n o w a = 1 b = 0$ $x^{2} + y^{2} = 1 \to c i r c l e$ $w h e n$ $a = 1 b = 1$ $x^{2} - 2 x^{2} y^{2} - 1 + x^{2} y^{2} + y^{2} = 0$ $x^{2} - x^{2} y^{2} - 1 + y^{2} = 0$ $x^{2} (1 - y^{2}) - (1 - y^{2}) = 0$ $(1 - y^{2}) (x^{2} - 1) = 0$ $(x^{2} - 1) (y^{2} - 1) = 0$ $a = 1 b = 2$ $x^{2} - 4 x^{2} y^{2} - 1 + 4 x^{2} y^{2} + y^{2} = 0$ $x^{2} + y^{2} = 1$ $a = 2 b = 2$ $4 x^{2} - 8 x^{2} y^{2} - 1 + 4 x^{2} y^{2} + y^{2} = 0$ $4 x^{2} - 4 x^{2} y^{2} - 1 + y^{2} = 0$ $4 x^{2} (1 - y^{2}) - (1 - y^{2}) = 0$ $(1 - y^{2}) (4 x^{2} - 1) = 0$ $(y^{2} - 1) (4 x^{2} - 1) = 0$ $p l s c h e c k . .$
	Commented by Tinkutara last updated on 03/Apr/19
	What is wrong in my method for B part?
	Commented by Tinkutara last updated on 03/Apr/19

	Commented by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

	$x^{2} {(y^{2} - 1)}^{2} = (1 - y^{2}) (1 - x^{2} y^{2})$ $x^{2} {(1 - y^{2})}^{2} = (1 - y^{2}) (1 - x^{2} y^{2})$ $x^{2} {(1 - y^{2})}^{2} - (1 - y^{2}) (1 - x^{2} y^{2}) = 0$ $(1 - y^{2}) (x^{2} - x^{2} y^{2} - 1 + x^{2} y^{2}) = 0$ $(1 - y^{2}) (x^{2} - 1) = 0$ $(x^{2} - 1) (y^{2} - 1) = 0$ $n o w y o u r s t e p s$ $x (y^{2} - 1) = \sqrt{(1 - y^{2}) (1 - x^{2} y^{2})}$ $x^{2} {(y^{2} - 1)}^{2} = (1 - y^{2}) (1 - x^{2} y^{2})$ $n x t s t e p s h o u l d b e ★ ★ b u t y o u c a n c e l l e d (y^{2} - 1) f r o m b o t h s i d e$ $x^{2} {(1 - y^{2})}^{2} - (1 - y^{2}) (1 - x^{2} y^{2}) = 0$ $s i n c e {(a - b)}^{2} = {(b - a)}^{2}$ $(1 - y^{2}) (x^{2} - x^{2} y^{2} - 1 + x^{2} y^{2}) = 0$ $(1 - y^{2}) (x^{2} - 1) = 0$ $(y^{2} - 1) (x^{2} - 1) = 0$
	Commented by Tinkutara last updated on 03/Apr/19
	Thank you so much Sir! I was confused with it for long!
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