Question

IIT JEE ADVANCE

If the two lines represented by
x²(tan²¢+cos²¢) - 2xytan¢ +y²sin²¢ = 0 make angle α,β with x axis then

(a) tanα+ tanβ = 4cosec2¢
(b) tanα tanβ= sec²¢ + tan²¢
(c) tanα -tanβ = 2
(d) tanα/tan β= 2+sin2¢ /2-sin2¢ ​

Answer 1

Answer:- (a,c,d)

Explanation:-

• Let the lines represented by the given Equation y = xtanα and y = xtanβ

given Equation ,x²(tan²¢ +cos²¢) -2xytan¢ + y²sin²¢ = 0

Dividing by sin²¢

we get ,

$y {}^{2} + \frac{2tan \theta \: xy}{sin { }^{2} \theta } + \frac{tan {}^{2} \theta \: + cos {}^{2} \theta}{sin {}^{2} \theta} x {}^{2} = 0$

∴ tanα + tanβ = 2tanθ/sin²θ

= 2tanθ/sin²θ => 2/sinθcosθ = 4cosec2θ ___(1)

∵ 2sinθcosθ = sin2θ

option (A) correct

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tanα tanβ = tan²θ + cos²θ /sin²θ

= sec²θ +cot²θ -----(2)

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now consider

(tanα - tanβ ) ² = (tanα + tanβ)² - 4tanα tan β

= 4/sin²¢ + cos² θ -4(sec²¢ +cot ²θ)using (1) and (2) equation

=4 [1-sin²¢ +cos⁴¢ / sin²¢cos²¢]

= 4[cos²¢ - cos⁴¢ / sin²¢ cos²¢

= 4 [ cos²¢ (1-cos²¢ )/sin²¢ cos²¢]

= 4[cos²¢ sin²¢ /cos²¢ sin²¢ ]

= 4

=> tanα - tanβ = +-2 ____(3)

Let us consider

tanα - tanβ = 2 _____(4)

Option (c) is correct

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then, on solving equation (1) and(4)

we get ' 2tanα = 4cosec2θ +2

and 2tanβ = 4cosec2θ -2;

then ' tanα /tanβ = 4cosec2θ +2/4cosec2θ -2

= 2+sin2θ/2-sin2θ

Option (D) is correct

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