The value of tan left( frac{1}{2} cos^{-1} le | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $\cos^{-1} \left( \frac{\sqrt{5}}{3} \right) = \alpha$. Then $\cos \alpha = \frac{\sqrt{5}}{3}$,where $0 < \alpha < \frac{\pi}{2}$. We need to

Vedclass · Accepted Answer

$\frac{3 - \sqrt{5}}{2}$

Vedclass · Answer

(C) Let $\cos^{-1} \left( \frac{\sqrt{5}}{3} ight) = \alpha$. Then $\cos \alpha = \frac{\sqrt{5}}{3}$,where $0 We need to find $	an \left( \frac{\alpha}{2} ight)$. Using the half-angle formula,$	an \left( \frac{\alpha}{2} ight) = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}$. Substituting the value of $\cos \alpha$: $	an \left( \frac{\alpha}{2} ight) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}$. To rationalize the denominator,multiply the numerator and denominator by $(3 - \sqrt{5})$: $\sqrt{\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}} = \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}} = \sqrt{\frac{(3 - \sqrt{5})^2}{4}} = \frac{3 - \sqrt{5}}{2}$.

The value of $\tan \left( \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right) \right)$ is

Similar Questions

Find the principal value of $\cot ^{-1}(\sqrt{3})$

$\sin^{-1}(\sin \frac{7\pi}{6}) = $

The solution of the inequality $(\cot^{-1}x)^2 - 5\cot^{-1}x + 6 > 0$ is:

Evaluate: $\sec^2(\tan^{-1} 2) + \csc^2(\cot^{-1} 3)$

Find the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module