If sin -1 x -sin -1 y -sin -1 z -3 -2- then xy - yx - xz -A- 0B- 1C- 1 or -1D- None

The correct option is B 1sin^ 1x + sin^ 1y+sin^ 1z =3 π/2 ⇒ sin^ 1x =sin^ 1y =sin^ 1z =π/2 [ ∵ π/2≤ sin^ 1x ≤π/2 ] ⇒ x=y=z=1 ⇒ xy+yz xz =1+1 1 =1 ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

If sin -1 x +...

Question

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}$ , then xy +yx -xz =

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Solution

The correct option is A 1
$s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2} \Rightarrow s i n^{- 1} x = s i n^{- 1} y = s i n^{- 1} z = \frac{π}{2} [∵ - \frac{π}{2} \leq s i n^{- 1} x \leq \frac{π}{2}]$
$\Rightarrow x = y = z = 1 \Rightarrow x y + y z - x z = 1 + 1 - 1 = 1$

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Q. If

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}

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Reason:

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If sin−1x+sin−1y+sin−1z=3π2, then xy +yx -xz =

The correct option is A 1sin−1x+sin−1y+sin−1z=3π2⇒sin−1x=sin−1y=sin−1z=π2[∵−π2≤sin−1x≤π2] ⇒x=y=z=1⇒xy+yz−xz=1+1−1=1

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}$ , then xy +yx -xz =

The correct option is A 1
$s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2} \Rightarrow s i n^{- 1} x = s i n^{- 1} y = s i n^{- 1} z = \frac{π}{2} [∵ - \frac{π}{2} \leq s i n^{- 1} x \leq \frac{π}{2}]$
$\Rightarrow x = y = z = 1 \Rightarrow x y + y z - x z = 1 + 1 - 1 = 1$