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QJEEAdv20Mae19b: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Mae19b Solution: Let x, y and z be positive real numbers. Suppose x, y and z are …

Let x, y and z be positive real numbers. Suppose x, y and z are the lengths of the sides of a triangle opposite to its angles X, Y, and Z, respectively. If

tan⁡X2+tan⁡Z2=2yx+y+z\tan {X \over 2} + \tan {Z \over 2} = {{2y} \over {x + y + z}} , then which of the following statements is/are TRUE?

Multiple Choice Options

  • A 2Y = X + Z
  • B Y = X + Z
  • C tan⁡X2\tan {X \over 2} = xy+z{x \over {y + z}}
  • D x2 + z2 −- y2 = xz

Step-by-Step Verified Solution

For a Δ\Delta XYZ, it is given that

tan⁡X2+tan⁡Z2=2yx+y+z\tan {X \over 2} + \tan {Z \over 2} = {{2y} \over {x + y + z}}

⇒Δs(s−x)+Δs(s−z)=ys\Rightarrow {\Delta \over {s(s - x)}} + {\Delta \over {s(s - z)}} = {y \over s}

⇒Δ(s−z+s−x)(s−x)(s−z)=y\Rightarrow \Delta {{(s - z + s - x)} \over {(s - x)(s - z)}} = y

⇒Δ=(s−x)(s−z)\Rightarrow \Delta = (s - x)(s - z)

⇒s(s−x)(s−y)(s−z)=(s−x)2(s−z)2\Rightarrow s(s - x)(s - y)(s - z) = {(s - x)^2}{(s - z)^2}

⇒s2−sy=s2−(x+z)s+xz\Rightarrow {s^2} - sy = {s^2} - (x + z)s + xz

⇒s(x+z−y)=xz\Rightarrow s(x + z - y) = xz

⇒(x+z)2−y2=2xz\Rightarrow {(x + z)^2} - {y^2} = 2xz

⇒x2+z2=y2⇒y=π2\Rightarrow {x^2} + {z^2} = {y^2} \Rightarrow y = {\pi \over 2}

∵\because X+Y+Z=π⇒X+Z=π2=YX + Y + Z = \pi \Rightarrow X + Z = {\pi \over 2} = Y

⇒X+Z=Y\Rightarrow X + Z = Y

∵\because tan⁡X2=1−cos⁡x1+cos⁡x=1−zy1+zy\tan {X \over 2} = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} = \sqrt {{{1 - {z \over y}} \over {1 + {z \over y}}}}

JEE Advanced 2020 Paper 1 Offline Mathematics - Properties of Triangle Question 5 English Explanation

=y−zy+z=y2−z2y+z= \sqrt {{{y - z} \over {y + z}}} = {{\sqrt {{y^2} - {z^2}} } \over {y + z}}

tan⁡X2=xy+z\tan {X \over 2} = {x \over {y + z}} { ∵\because x2 + z2 = y2}

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