Question

If $\sin^{-1}x +\sin^{-1}y+\sin^{-1}z=\pi $ , then $x^4+y^4+z^4+4x^2\,y^2\,z^2=K(x^2\,y^2+y^2\,z^2+z^2\,x^2)$ , where K =

• A. 1
• B. 2
• C. 4
• D. none of these.

Answer 1

Answer: (B)

Since $sin^{-1}x+sin^{-1} y = \pi-sin^{-1} z = \pi$
$ \therefore sin^{-1} x +sin^{-1} y = \pi - sin^{-1}z$
$ \Rightarrow sin^{-1}\left[x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right]$
$ = \pi -sin^{-1}\left(z\right) $
$ \Rightarrow x \sqrt{1-y^{2}} + y \sqrt{1-x^{2}}$
$= sin \left(\pi-sin^{-1} \left(z\right)\right) $
$= sin \left(sin^{-1} z\right) = z$
$ \Rightarrow x^{2} \left(1-y^{2}\right) = z^{2} +y^{2}\left(1-x^{2}\right)-2zy \sqrt{1-x^{2}} $
$\Rightarrow \left(x^{2}+y^{2}-z^{2}\right) = 4y^{2}z^{2}\left(1-x^{2}\right)$
$\Rightarrow x^{4}+y^{4}+z^{4}-2x^{2}y^{2} -2x^{2}z^{2} +2y^{2}z^{2} $
$= 4y^{2}z^{2}-4x^{2}y^{2}z^{2} $
$\Rightarrow x^{4}+y^{4}+z^{4} +4x^{2}y^{2}z^{2} $
$ = 2\left(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\right) $
$ \therefore K = 2$