If P and x are real- then e 2 - -1 x 1- ix - ix -1 p -A- P B- e ixC- 1D- None

The correct option is C 1 Let cot^ 1x = θ.Then tanθ = 1/x ∴ e^2i cot^ 1x = e^2iθ = cos2θ + i sin2θ Let = 1 tan^2θ/1 + tan^2θ + i 2tanθ/1 + tan^2θ = x^2 1/ ...

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Standard XII

Mathematics

Inverse Function

If P and x ar...

Question

If P and x are real, then $e^{2 π c o t^{- 1} x} {(\frac{1 + i x}{i x - 1})}^{p} =$

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Solution

The correct option is A
1

Let $c o t^{- 1} x = θ$ .Then tan $θ = \frac{1}{x}$

$∴ e^{2 i c o t^{- 1} x} = e^{2 i θ} = c o s 2 θ + i s i n 2 θ$

Let = $\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ} + i \frac{2 t a n θ}{1 + t a n^{2} θ} = \frac{x^{2} - 1}{x^{2} + 1} + \frac{2 x}{1 + x^{2}} i$

= $\frac{(x + i)^{2}}{(x + i) (x - i)} = \frac{x + i}{x - i} = \frac{i x - 1}{i x + 1}$

$∴ e^{2 p i c o t^{- 1} x} {(\frac{i x + 1}{i x - 1})}^{p} = {(\frac{i x - 1}{i x + 1})}^{p} {(\frac{i x + 1}{i x - 1})}^{p} = 1$

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If P and x are real, then e2πcot−1x(1+ixix−1)p=

The correct option is A 1 Let cot−1x=θ.Then tanθ=1x ∴e2icot−1x=e2iθ=cos2θ+isin2θ Let = 1−tan2θ1+tan2θ+i2tanθ1+tan2θ=x2−1x2+1+2x1+x2i = (x+i)2(x+i)(x−i)=x+ix−i=ix−1ix+1 ∴e2picot−1x(ix+1ix−1)p=(ix−1ix+1)p(ix+1ix−1)p=1

If P and x are real, then $e^{2 π c o t^{- 1} x} {(\frac{1 + i x}{i x - 1})}^{p} =$