$\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$ has the value

$\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$ has the value
  1. $\frac{\sqrt{5}}{6}$
  2. $\frac{\sqrt{5}}{6}$
  3. $\frac{3-\sqrt{5}}{2}$
  4. $\frac{3+\sqrt{5}}{2}$

Solution

Let $\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}=\theta$ $\cos 2 \theta=\frac{\sqrt{5}}{3}$ Now $\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$ $\begin{aligned} & =\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} \\ & =\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\frac{3-\sqrt{5}}{2} \end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

Practice more Trigonometric Equations questions on Aicharya