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Mathematics Question on Inverse Trigonometric Functions

If sin⁡−1x+sin⁡−1y+sin⁡−1z=π\sin^{-1}x +\sin^{-1}y+\sin^{-1}z=\pi , then x4+y4+z4+4x2 y2 z2=K(x2 y2+y2 z2+z2 x2)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

2

Explanation

Solution

Since sin−1x+sin−1y=π−sin−1z=πsin^{-1}x+sin^{-1} y = \pi-sin^{-1} z = \pi ∴sin−1x+sin−1y=π−sin−1z \therefore sin^{-1} x +sin^{-1} y = \pi - sin^{-1}z ⇒sin−1[x1−y2+y1−x2] \Rightarrow sin^{-1}\left[x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right] =π−sin−1(z)= \pi -sin^{-1}\left(z\right) ⇒x1−y2+y1−x2 \Rightarrow x \sqrt{1-y^{2}} + y \sqrt{1-x^{2}} =sin(π−sin−1(z))= sin \left(\pi-sin^{-1} \left(z\right)\right) =sin(sin−1z)=z= sin \left(sin^{-1} z\right) = z ⇒x2(1−y2)=z2+y2(1−x2)−2zy1−x2\Rightarrow x^{2} \left(1-y^{2}\right) = z^{2} +y^{2}\left(1-x^{2}\right)-2zy \sqrt{1-x^{2}} ⇒(x2+y2−z2)=4y2z2(1−x2)\Rightarrow \left(x^{2}+y^{2}-z^{2}\right) = 4y^{2}z^{2}\left(1-x^{2}\right) ⇒x4+y4+z4−2x2y2−2x2z2+2y2z2\Rightarrow x^{4}+y^{4}+z^{4}-2x^{2}y^{2} -2x^{2}z^{2} +2y^{2}z^{2} =4y2z2−4x2y2z2= 4y^{2}z^{2}-4x^{2}y^{2}z^{2} ⇒x4+y4+z4+4x2y2z2\Rightarrow x^{4}+y^{4}+z^{4} +4x^{2}y^{2}z^{2} =2(x2y2+y2z2+z2x2)= 2\left(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\right) ∴K=2 \therefore K = 2