If sin -1 x -sin -1 y -sin -1 z -2- then x 2- y 2- z 2-2 xyz -A- 0B- 1C- -1D- None

The correct option is B 1Given expression ⇒ cos^ 1√(1 x^2)+cos^ 1√(1 y^2)=π/2 sin^ 1z ⇒ cos^ 1 [ √((1 x^2)(1 y^2)) √({ 1 (1 x^2) }{ 1 (1 y^2) }) ]=cos^ 1z ⇒ ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If sin -1 x +...

Question

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{π}{2}, t h e n x^{2} + y^{2} + z^{2} + 2 x y z =$

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Solution

The correct option is B 1
Given expression
$\Rightarrow c o s^{- 1} \sqrt{1 - x^{2}} + c o s^{- 1} \sqrt{1 - y^{2}} = \frac{π}{2} - s i n^{- 1} z$
$\Rightarrow$ $c o s^{- 1} [\sqrt{(1 - x^{2}) (1 - y^{2})} - \sqrt{{1 - (1 - x^{2})} {1 - (1 - y^{2})}}] = c o s^{- 1} z$
$\Rightarrow$ $\sqrt{(1 - x^{2}) (1 - y^{2})} - x y = z$
$\Rightarrow$ $(1 - x^{2}) (1 - y^{2}) = (x y + z)^{2}$
$\Rightarrow$ $x^{2} + y^{2} + z^{2} + 2 x y z = 1$

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If sin−1x+sin−1y+sin−1z=π2,thenx2+y2+z2+2xyz=

The correct option is B 1Given expression ⇒cos−1√1−x2+cos−1√1−y2=π2−sin−1z ⇒ cos−1[√(1−x2)(1−y2)−√{1−(1−x2)}{1−(1−y2)}]=cos−1z ⇒ √(1−x2)(1−y2)−xy=z ⇒ (1−x2)(1−y2)=(xy+z)2 ⇒ x2+y2+z2+2xyz=1

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{π}{2}, t h e n x^{2} + y^{2} + z^{2} + 2 x y z =$