First slide

Properties of ITF

Question

$S e c^{- 1} (\sqrt{1 + x^{2}}) + C o s e c e^{- 1} (\frac{\sqrt{1 + y^{2}}}{y}) + C o t^{- 1} (\frac{1}{z}) = π$ then

Moderate

A

$x + y + z = 0$
B

$x + y + z = 1$
C

$x + y + z = x y z$
D

$x + y + z = - x y z$

Solution

$L e t x = \tan A, y = \tan B a n d z = \tan C . T h e n s e c^{- 1} \sqrt{1 + x^{2}} + \cos e c^{- 1} (\frac{\sqrt{1 + y^{2}}}{y}) + c o t^{- 1} \frac{1}{z} = π \Rightarrow s e c^{- 1} \sqrt{1 + \tan^{2} A} + \cos e c^{- 1} (\frac{\sqrt{1 + \tan^{2} B}}{\tan B}) + c o t^{- 1} \frac{1}{\tan C} = π \Rightarrow s e c^{- 1} (s e c A) + \cos e c^{- 1} (\cos e c B) + c o t^{- 1} (c o t C) = π \Rightarrow A + B + C = π \Rightarrow tanA + tanB + tanC = tanAtanBtanC \Rightarrow x + y + z = xyz$

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