Sin -1 pi x -tan -1 frac x p i lsin -1 frac -x p i - -1 pi x lendbmatrix - -cos -1 pi x -tan -1frac x p i lsin-1 frac x p i -tan -1 pi x lend bmatrix then A - B is equal toA- IB- 0C- 2 ID- 1-2 I

The correct option is D 1/2 IA B=1/π[ π/2 amp;0; 0 amp;π/2; ] A B=1/2I ...

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Evaluation of a Determinant

sin ∧-1 pi x ...

Question

If $A = \frac{1}{π} [\begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & {cot}^{- 1} (π x) \end{matrix}],$
$B = \frac{1}{π} [\begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & - {tan}^{- 1} (π x) \end{matrix}]$
then $A - B$ is equal to

I

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0.0

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2 I

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\frac{1}{2} I

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A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

A - B

A = \frac{1}{π} ⎡ ⎢ ⎢ ⎣ \begin{matrix} s i n^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦, B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} - c o s^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & t a n^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦

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