If cos(cot−1(12))=cot(cos−1x), then a value of4x is

Cot−1(12)=θ⇒Cotθ=12Cosθ=15∴ Cot(Cos−1x)=15Let Cos−1(x)=α⇒cosα=xCotα=x1−x2∴ x1−x2=15x=16

If cos(cot−1(12))=cot(cos−1x), then a value of4x is

Cot−1(12)=θ⇒Cotθ=12Cosθ=15∴ Cot(Cos−1x)=15Let Cos−1(x)=α⇒cosα=xCotα=x1−x2∴ x1−x2=15x=16

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If $\cos (\cot^{- 1} (\frac{1}{2})) = \cot (\cos^{- 1} x),$ then a value of $4 x$ is

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$\frac{4}{\sqrt{6}}$

$\frac{- 1}{\sqrt{12}}$

$\frac{2}{\sqrt{6}}$

$\frac{- 2}{\sqrt{6}}$

answer is A.

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$C o t^{- 1} (\frac{1}{2}) = θ \Rightarrow C o t θ = \frac{1}{2}$

$C o s θ = \frac{1}{\sqrt{5}}$

$∴ C o t (C o s^{- 1} x) = \frac{1}{\sqrt{5}}$

Let $C o s^{- 1} (x) = α \Rightarrow \cos α = x$

$C o t α = \frac{x}{\sqrt{1 - x^{2}}}$

$∴ \frac{x}{\sqrt{1 - x^{2}}} = \frac{1}{\sqrt{5}}$

$x = \frac{1}{\sqrt{6}}$

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If cos(cot−1(12))=cot(cos−1x), then a value of4x is

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If $\cos (\cot^{- 1} (\frac{1}{2})) = \cot (\cos^{- 1} x),$ then a value of $4 x$ is