If $cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi,$ then :

Question

Accepted Answer

To determine why Option 1 is the correct answer, we need to analyze the given condition and the properties of inverse trigonometric functions.

### Explanation for Option 1:

The condition given is $ \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi $. This implies that the angles whose cosines are $ x, y, $ and $ z $ respectively, sum up to $ \pi $. In trigonometry, if three angles $ A, B, $ and $ C $ satisfy $ A + B + C = \pi $, then they can be considered as the angles of a triangle.

For angles $ A, B, $ and $ C $ in a triangle, the cosine rule states that:

$$ \cos A + \cos B + \cos C = 1 + \frac{r}{R} $$

where $ r $ is the inradius and $ R $ is the circumradius of the triangle. However, in this context, we are dealing with the identity related to the sum of angles in a triangle:

$$ \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1 $$

This identity directly translates to the given variables $ x, y, $ and $ z $ as:

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

Thus, Option 1 is correct because it directly follows from the identity involving the cosines of angles in a triangle.

### Explanation for why other options are incorrect:

- **Option 2:** $ \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \cos^{-1}x + \cos^{-1}y + \cos^{-1}z $

This option is incorrect because the sum of inverse sine functions does not equal the sum of inverse cosine functions unless specific conditions are met, which are not generally true for arbitrary $ x, y, $ and $ z $.

- **Option 3:** $ xy + yz + zx = x + y + z - 1 $

This equation does not hold true under the given condition $ \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi $. There is no trigonometric identity or geometric interpretation that supports this equation as a consequence of the given condition.

- **Option 4:** $ (x+\frac{1}{x})+(y+\frac{1}{y})+(z+\frac{1}{z}) \ge 6 $

This inequality is not generally true for all $ x, y, $ and $ z $ satisfying the given condition. The expression $ x + \frac{1}{x} $ is minimized at 2 when $ x = 1 $, but there is no guarantee that all terms will sum to 6 or more without additional constraints on $ x, y, $ and $ z $.

In conclusion, Option 1 is the correct answer because it is derived from a well-known trigonometric identity involving the cosines of angles in a triangle, which directly applies to the given condition.