If \(\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot…

If \(\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot \left(\frac{\pi}{24}\right)\right)=\sqrt{a^2+a}+\sqrt{a}\), then \(a=\)
  1. 3
  2. 2
  3. 1
  4. 4

Solution

To get \(\cot \frac{\pi}{24}\), we proceed as follows \(\begin{aligned} \cot \frac{\pi}{24} & =\frac{2 \cos ^2 \frac{\pi}{24}}{2 \sin \frac{\pi}{24} \cdot \cos \frac{\pi}{24}} \\ & =\frac{1+\cos \frac{\pi}{12}}{\sin \frac{\pi}{12}}=\frac{1+\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)}{1+\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)} \end{aligned}\) \(\begin{aligned} & =\frac{1+\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}}{\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}} \\ & =2+\sqrt{2}+\sqrt{3}+\sqrt{6} \\ & \text {and } \tan \frac{\pi}{24}=-2+\sqrt{2}-\sqrt{3}+\sqrt{6} \\ & \text {So, } \frac{1}{2}\left(\tan \frac{\pi}{24}+\cot \frac{\pi}{24}\right)=\sqrt{2}+\sqrt{6}=\sqrt{2^2+2}+\sqrt{2} \\ & =\sqrt{a^2+a}+\sqrt{a} \quad \text { (given) } \\ \end{aligned}\) So, \(a=2\)

Asked in: AP EAMCET 2020 (17 Sep Shift 2)

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