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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2017 · Mathematics · Inverse Trigonometric Functions

Let x, y, z be positive real numbers such that x, y, z are in GP...

Let x,y,z\mathrm{x}, \mathrm{y}, \mathrm{z} be positive real numbers such that x,y,z\mathrm{x}, \mathrm{y}, \mathrm{z} are in GP and tan⁡−1x,tan⁡−1y\tan ^{-1} x, \tan ^{-1} y and tan⁡−1z\tan ^{-1} z are in AP. Then which one of the following is correct?

Options

  1. Ax=y=z\mathrm{x}=\mathrm{y}=\mathrm{z}
  2. Bxz=1\mathrm{xz}=1
  3. Cx≠yx \neq y and y=zy=z
  4. Dx=yx=y and y≠zy \neq z

Correct answer

A. x=y=z\mathrm{x}=\mathrm{y}=\mathrm{z}

Step-by-step solution

2tan⁡−1y=tan⁡−1x+tan⁡−1z(∴tan⁡−1,tan⁡−1y,tan⁡−1z are in  A.P ) and x,y,z are in G.P ∴y2=xz...(i)\begin{aligned} & 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} z \\ & \left(\therefore \tan ^{-1}, \tan ^{-1} y, \tan ^{-1} z \text{ are\text{ in } A.P }\right) \\ & \text{ and } x, y, z \text{ are in G.P } \quad \therefore \quad y^2=x z...(i) \end{aligned}⇒2y1−y2=x+z1−xz⇒2y1−y2=x+z1−y2⇒2y=x+zx,y,z are in A.P. \begin{aligned} & \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \\ & \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-y^2} \\ & \Rightarrow 2 y=x+z \\ & x, y, z \text{ are in A.P. } \end{aligned}(from(i))Given(from (i)) Givenx, y, zarealsoinG.PSo,are also in G.P So,x=y=z$
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