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