We are asked to find the value of \(\tan\left(\frac{1}{2}\cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right)\). To solve this, we'll use the half-angle identity for tangent:
The half-angle formula for tangent is:
\(\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}}\)
Let's set \(\cos(\theta) = \frac{\sqrt{5}}{3}\). This means \(\theta = \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\).
Plugging this into the half-angle formula, we have:
\(\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}}\)
Simplifying the expression inside the square root:
This gives us:
\(\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{\frac{3 + \sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}\)
Next, to simplify \(\frac{3 - \sqrt{5}}{3 + \sqrt{5}}\), we multiply by the conjugate:
\( \frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{(3 - \sqrt{5})^2}{3^2 - (\sqrt{5})^2} = \frac{9 - 6\sqrt{5} + 5}{9 - 5} = \frac{14 - 6\sqrt{5}}{4}\)
Therefore:
\(\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{14 - 6\sqrt{5}}{4}} = \frac{\sqrt{14 - 6\sqrt{5}}}{2}\)
This further simplifies to:
\(\tan\left(\frac{\theta}{2}\right) = \frac{1}{2}(3 - \sqrt{5})\)
Thus, the correct answer is option: \(\frac{1}{2}(3-\sqrt{5})\).
Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>, \theta \in(0,2 \pi)$, holds If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :