If y = log_e tan left(frac{pi}{4} + frac{x}{2 | vedclass

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

Question

(D) Given $y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$.$e^y = 	an \left(\frac{\pi}{4} + \frac{x}{2}ight) = \frac{1 + 	an \frac{x}{2}

Vedclass · Accepted Answer

$	an \frac{x}{2}$

Vedclass · Answer

(D) Given $y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$.$e^y = 	an \left(\frac{\pi}{4} + \frac{x}{2}ight) = \frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}}$.We know that $	anh \left(\frac{y}{2}ight) = \frac{e^y - 1}{e^y + 1}$.Substituting the value of $e^y$:$	anh \left(\frac{y}{2}ight) = \frac{\frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} - 1}{\frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} + 1} = \frac{(1 + 	an \frac{x}{2}) - (1 - 	an \frac{x}{2})}{(1 + 	an \frac{x}{2}) + (1 - 	an \frac{x}{2})} = \frac{2 	an \frac{x}{2}}{2} = 	an \frac{x}{2}$.Thus,option $D$ is correct.

If $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$,then $\tanh \left(\frac{y}{2}\right) = $

Similar Questions

Find the value of $\sin \frac{31 \pi}{3}$.

If $\sin A = \frac{-7}{25}$,$\cos B = \frac{8}{17}$,$A$ does not lie in the $3^{\text{rd}}$ quadrant and $B$ does not lie in the $1^{\text{st}}$ quadrant,then $8 \tan A - 5 \cot B =$

$\frac{\sin \theta}{1-\cot \theta} + \frac{\cos \theta}{1-\tan \theta} = $

If $\operatorname{cosech} x = \frac{4}{5}$,then $\cosh x =$

The value of $\sin 18^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module