If theta = frac{pi}{12} and x = log left(cot | vedclass

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

Question

(A) Given: $x = \log \left(\cot \left(\frac{\pi}{4} + 	hetaight)ight)$Substitute $	heta = \frac{\pi}{12}$:$x = \log \left(\cot \left(\frac{\pi}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given: $x = \log \left(\cot \left(\frac{\pi}{4} + 	hetaight)ight)$Substitute $	heta = \frac{\pi}{12}$:$x = \log \left(\cot \left(\frac{\pi}{4} + \frac{\pi}{12}ight)ight) = \log \left(\cot \left(\frac{3\pi + \pi}{12}ight)ight) = \log \left(\cot \left(\frac{\pi}{3}ight)ight)$Since $\cot \left(\frac{\pi}{3}ight) = \frac{1}{\sqrt{3}}$,we have $x = \log \left(\frac{1}{\sqrt{3}}ight) = -\log \sqrt{3} = -\frac{1}{2} \log 3$.Then $e^x = \frac{1}{\sqrt{3}}$ and $e^{-x} = \sqrt{3}$.The definition of $\cosh x$ is $\frac{e^x + e^{-x}}{2}$.$\cosh x = \frac{\frac{1}{\sqrt{3}} + \sqrt{3}}{2} = \frac{\frac{1 + 3}{\sqrt{3}}}{2} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.

If $\theta = \frac{\pi}{12}$ and $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$,then $\cosh x =$

Similar Questions

$\sqrt{3 + \sqrt{5}} - \sqrt{2 + \sqrt{3}} = $

What is the cube root of $9\sqrt{3} + 11\sqrt{2}$?

If $\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$,then $x=$

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

Evaluate: $\frac{\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} - 1}{\sqrt{5 + 2\sqrt{6}}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module