First slide

Introduction to ITF

Question

$If \tan^{- 1} (\tan \frac{5 π}{4}) = α, \tan^{- 1} (- \tan \frac{2 π}{3}) = β then$

Easy

A

$4 α - 4 β = 0$
B

$4 α - 3 β = 0$
C

$α > β$
D

None of these

Solution

$Clearly, \tan^{- 1} [\tan \frac{5 π}{4}] = \tan^{- 1} [\tan (π + \frac{π}{4})] = \tan^{- 1} (\tan \frac{π}{4}) = \frac{π}{4} \Rightarrow α = \frac{π}{4} and \tan^{- 1} [- \tan (\frac{2 π}{3})] = \tan^{- 1} [- \tan (π - \frac{π}{3})] = \tan^{- 1} (\tan \frac{π}{3}) = \frac{π}{3} \Rightarrow β = \frac{π}{3} Now, consider 4 α - 3 β = 4 (\frac{π}{4}) - 3 (\frac{π}{3}) = 0$

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