Suppose that tanalpha =3 and 0^circ lt alpha lt 90^circ Find the exact values of cos(alpha )/(2) and tan(alpha )/(2) cos(alpha )/(2)= tan(alpha )/(2)= square

To find the exact values of $\cos\frac{\alpha}{2}$ and $\tan\frac{\alpha}{2}$, we can use the half-angle formulas.<br /><br />Given that $\tan\alpha = 3$ and $0^{\circ} < \alpha < 90^{\circ}$, we can find the value of $\sin\alpha$ and $\cos\alpha$ using the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$.<br /><br />Since $\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = 3$, we can let $\sin\alpha = 3k$ and $\cos\alpha = k$. Substituting these values into the Pythagorean identity, we get:<br /><br />$(3k)^2 + k^2 = 1$<br /><br />$9k^2 + k^2 = 1$<br /><br />$10k^2 = 1$<br /><br />$k^2 = \frac{1}{10}$<br /><br />$k = \frac{1}{\sqrt{10}}$<br /><br />Therefore, $\sin\alpha = 3k = \frac{3}{\sqrt{10}}$ and $\cos\alpha = k = \frac{1}{\sqrt{10}}$.<br /><br />Now, let's find $\cos\frac{\alpha}{2}$ and $\tan\frac{\alpha}{2}$ using the half-angle formulas.<br /><br />The half-angle formula for cosine is:<br /><br />$\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}$<br /><br />Since $0^{\circ} < \alpha < 90^{\circ}$, we know that $0^{\circ} < \frac{\alpha}{2} < 45^{\circ}$. Therefore, $\cos\frac{\alpha}{2}$ is positive.<br /><br />Substituting the value of $\cos\alpha$, we get:<br /><br />$\cos\frac{\alpha}{2} = \sqrt{\frac{1 + \frac{1}{\sqrt{10}}}{2}}$<br /><br />Simplifying further, we have:<br /><br />$\cos\frac{\alpha}{2} = \sqrt{\frac{\sqrt{10} + 1}{2\sqrt{10}}}$<br /><br />The half-angle formula for tangent is:<br /><br />$\tan\frac{\alpha}{2} = \pm\frac{1 - \cos\alpha}{\sin\alpha}$<br /><br />Again, since $0^{\circ} < \alpha < 90^{\circ}$, we know that $0^{\circ} < \frac{\alpha}{2} < 45^{\circ}$. Therefore, $\tan\frac{\alpha}{2}$ is positive.<br /><br />Substituting the values of $\sin\alpha$ and $\cos\alpha$, we get:<br /><br />$\tan\frac{\alpha}{2} = \frac{1 - \frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}}$<br /><br />Simplifying further, we have:<br /><br />$\tan\frac{\alpha}{2} = \frac{\sqrt{10} - 1}{3}$<br /><br />Therefore, the exact values of $\cos\frac{\alpha}{2}$ and $\tan\frac{\alpha}{2}$ are:<br /><br />$\cos\frac{\alpha}{2} = \sqrt{\frac{\sqrt{10} + 1}{2\sqrt{10}}}$<br /><br />$\tan\frac{\alpha}{2} = \frac{\sqrt{10} - 1}{3}$