If A - 1- sin-1- x tan-1x- - sin-1 x- cot-1- x - B - - -cos-1- x tan-1x- - sin-1 x- tan-1- x - then A - B is equal to

The correct option is D 12I A= 1 π #160; [ sin ^ 1 ( π x ) #160; #160; amp; tan ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

If A = 1π[ ...

Question

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

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\frac{1}{2} I

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Solution

The correct option is D $\frac{1}{2} I$
$A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & {cot}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$A - B = \frac{1}{π} [\begin{matrix} {sin}^{- 1} (π x) + {cos}^{- 1} (π x) & 0 0 & {cot}^{- 1} (π x) + {tan}^{- 1} (π x) \end{matrix}]$
$A - B = \frac{1}{π} [\begin{matrix} π / 2 & 0 0 & π / 2 \end{matrix}]$
$A - B = \frac{1}{π} * \frac{π}{2} [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$
$A - B = \frac{1}{2} I$

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Similar questions

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{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

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If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ ,then $c o s^{- 1} x + c o s^{- 1} y$ is equal to

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If A=1π⎡⎢ ⎢⎣sin−1(πx)tan−1(xπ)sin−1(xπ)cot−1(πx)⎤⎥ ⎥⎦,B=⎡⎢ ⎢⎣−cos−1(πx)tan−1(xπ)sin−1(xπ)tan−1(πx)⎤⎥ ⎥⎦ then A - B is equal to