If tan-11-x-1-x - 1-2 tan-1x- then x -

We have tan^ 11 x/1+x = 1/2 tan^ 1x ⇒ tan^ 1 [ 1 tan θ/1+tan θ ] = 1/2θ (Putting x = tan θ) ⇒ tan^ 1 [ tan π/4 tanθ/1+tan π/4tanθ ] = θ/2 ⇒ tan^ 1 tan ( π/4 θ ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If tan-11-x/1...

Question

If $t a n^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} t a n^{- 1} x,$ then x =

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\sqrt{3}

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\frac{1}{\sqrt{3}}

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None of these

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Solution

The correct option is C $\frac{1}{\sqrt{3}}$
we have $t a n^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} t a n^{- 1} x$
$\Rightarrow t a n^{- 1} [\frac{1 - t a n θ}{1 + t a n θ}] = \frac{1}{2} θ$ (Putting $x = t a n θ$ )
$\Rightarrow t a n^{- 1} [\frac{t a n \frac{π}{4} - t a n θ}{1 + t a n \frac{π}{4} t a n θ}] = \frac{θ}{2} \Rightarrow t a n^{- 1} t a n (\frac{π}{4} - θ) = \frac{θ}{4} - θ = \frac{θ}{2} \Rightarrow θ = \frac{π}{6} = t a n^{- 1} x \Rightarrow x = t a n \frac{π}{6} = \frac{1}{\sqrt{3}} .$

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If tan−11−x1+x=12 tan−1x, then x =

The correct option is C 1√3we have tan−11−x1+x=12 tan−1x ⇒tan−1[1−tan θ1+tan θ]=12θ (Putting x=tanθ) ⇒tan−1[tanπ4−tanθ1+tanπ4tanθ]=θ2⇒tan−1 tan(π4−θ)=θ4−θ=θ2⇒θ=π6=tan−1 x⇒x=tanπ6=1√3.

If $t a n^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} t a n^{- 1} x,$ then x =