Question

Sin inverse X + Sin inverse Y + Sin inverse Z is equal to pi by 2 then find X square + Y square + Z square + 2 x y z =?

Answer 1

Answer:

$x^2+y^2+z^2+2xyz=1$

Step-by-step explanation:

Formula used:

$1.\:sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$

$2.\:cos\theta=\sqrt{1-sin^2\theta}$

Given:

$sin^{-1}x+sin^{-1}y+sin^{-1}z=\frac{\pi}{2}$

$sin^{-1}x+sin^{-1}y=\frac{\pi}{2}-sin^{-1}z$

$sin^{-1}x+sin^{-1}y=cos^{-1}z$............(1)

Take

$sin^{-1}x=A$

$sinA=x$

$cosA=\sqrt{1-sin^2A}=\sqrt{1-x^2}$

$sin^{-1}y=B$

$sinB=y$

$cosB=\sqrt{1-sin^2B}=\sqrt{1-y^2}$

Now (1) becomes

$A+B=cos^{-1}z$

$cos(A+B)=z$

$cosA\:cosB-sinA\:sinB=z$

$\sqrt{1-x^2}\:\sqrt{1-y^2}-xy=z$

$\sqrt{1-x^2}\:\sqrt{1-y^2}=xy+z$

squaring on both sides we get

$(1-x^2)(1-y^2)=(xy+z)^2$

$1-x^2-y^2+x^2y^2=x^2y^2+z^2+2xyz$

$1-x^2-y^2=z^2+2xyz$

Rearranging terms we get

$x^2+y^2+z^2+2xyz=1$