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Question: If \(x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 2 x y z\)is equal to....

If x2+y2+z2+2xyzx ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 2 x y zis equal to.

A

0

B

1

C

2

D

3

Answer

1

Explanation

Solution

sin⁡−1x+sin⁡−1y+sin⁡−1z=π2\sin ^ { - 1 } x + \sin ^ { - 1 } y + \sin ^ { - 1 } z = \frac { \pi } { 2 }

Put sin⁡−1x=α\sin ^ { - 1 } x = \alpha, sin⁡−1y=β\sin ^ { - 1 } y = \beta sin⁡−1z=γ\sin ^ { - 1 } z = \gamma

α+β+γ=π2\alpha + \beta + \gamma = \frac { \pi } { 2 }, (Given)

or α+β=π2−γ\alpha + \beta = \frac { \pi } { 2 } - \gamma or cos⁡(α+β)=cos⁡(π2−γ)\cos ( \alpha + \beta ) = \cos \left( \frac { \pi } { 2 } - \gamma \right)

cos⁡αcos⁡β−sin⁡αsin⁡β=sin⁡γ\cos \alpha \cos \beta - \sin \alpha \sin \beta = \sin \gamma …..(i)

and, we have sin⁡α=x\sin \alpha = x ⇒ cos⁡α=1−x2\cos \alpha = \sqrt { 1 - x ^ { 2 } }

Similarly, cos⁡β=1−y2\cos \beta = \sqrt { 1 - y ^ { 2 } }

∴\therefore From equation (i), we get 1−x2⋅1−y2=xy+z\sqrt { 1 - x ^ { 2 } } \cdot \sqrt { 1 - y ^ { 2 } } = x y + z

Squaring both sides, we have x2+y2+z2+2xyz=1x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 2 x y z = 1 .