If A = 1 [ sin - 1 ( x ) tan - 1 ( x ) sin - 1 ( x ) cot - 1 ( x ) ] , B = 1 [ - cos - 1 ( x ) tan - 1 ( x ) sin - 1 ( x ) - tan - 1 ( x ) ] , then A - B is equal to:

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:

Smart Achievers · Accepted Answer

(4)

Ans. $\frac{1}{2} I$

Explanation:

We have,

$B = [\begin{matrix} - \frac{1}{π} \cos^{- 1} (π x) & \frac{1}{π} \tan^{- 1} (\frac{x}{π}) \\ \frac{1}{π} \sin^{- 1} (\frac{x}{π}) & - \frac{1}{π} \tan^{- 1} (π x) \end{matrix}]$

and $A = [\begin{matrix} \frac{1}{π} \sin^{- 1} (π x) & \frac{1}{π} \tan^{- 1} (\frac{x}{π}) \\ \frac{1}{π} \sin^{- 1} (\frac{x}{π}) & \frac{1}{π} \cot^{- 1} (π x) \end{matrix}]$

$∴$ $A - B = [\begin{matrix} \frac{1}{π} (\sin^{- 1} (π x) + \cos^{- 1} (π x)) & 0 \\ 0 & \frac{1}{π} (\cot^{- 1} (π x) + \tan^{- 1} (π x)) \end{matrix}]$

$= [\begin{matrix} \frac{1}{π} \cdot \frac{π}{2} & 0 \\ 0 & \frac{1}{π} \cdot \frac{π}{2} \end{matrix}] = \frac{1}{2} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = \frac{1}{2} I$

Solution

1 $I$

2 $O$

3 $2 I$

4 $\frac{1}{2} I$

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Solution

Q. If A=1π[sin-1(πx)tan-1(xπ)sin-1(xπ)cot-1(πx)], B=1π[-cos-1(πx)tan-1(xπ)sin-1(xπ)-tan-1(πx)], then A-B is equal to:

1 I

2 O

3 2I

4 12I

Ans. (4) Ans. 12I Explanation: We have, B=[-1πcos-1(πx)1πtan-1(xπ)1πsin-1(xπ)-1πtan-1(πx)] and A=[1πsin-1(πx)1πtan-1(xπ)1πsin-1(xπ)1πcot-1(πx)] ∴ A-B=[1π(sin-1(πx)+cos-1(πx))001π(cot-1(πx)+tan-1(πx))] =[1π·π2001π·π2]=12[1001]=12I

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1 $I$

2 $O$

3 $2 I$

4 $\frac{1}{2} I$