Mathematics

# $\displaystyle\dfrac{e^{2x}-1}{e^{2x}+1}$ in terms of hyperbolic function is

##### SOLUTION
$\dfrac{{e}^{2x}-1}{{e}^{2x}+1}$
Remove ${e}^{x}$ from both the numerator and denominator.
$=\dfrac{{e}^{x}\left({e}^{x}-{e}^{-x}\right)}{{e}^{x}\left({e}^{x}+{e}^{-x}\right)}$
$=\dfrac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$
$\tan{hx}$ which is a hyperbolic function

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Hard
Evaluate:$\displaystyle \int \dfrac{dx}{2+2\sin{2x}}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Solve $\displaystyle\int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx$.
• A. $I =\sqrt{2}{{\tan }^{-1}}\left( \dfrac{1+\tan x}{\sqrt{2\tan x}} \right)+{{C}_{1}}$
• B. $I ={{\tan }^{-1}}\left( \dfrac{1-\tan x}{\sqrt{2\tan x}} \right)+{{C}_{1}}$
• C. None of these
• D. $I =\sqrt{2}{{\tan }^{-1}}\left( \dfrac{1-\tan x}{\sqrt{2\tan x}} \right)+{{C}_{1}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate $\displaystyle \int_{0}^{2\pi}\frac{\sin ^{2}\theta }{a-b \cos \theta }d\theta$ for $a> b> 0$
• A. $\dfrac{2\pi }{a^{2}}\left [ -a+\sqrt{\left ( a^{2}+b^{2} \right )} \right ].$
• B. $\dfrac{2\pi }{a^{2}}\left [ a+\sqrt{\left ( a^{2}-b^{2} \right )} \right ].$
• C. $\dfrac{2\pi }{b^{2}}\left [ a+\sqrt{\left ( a^{2}+b^{2} \right )} \right ].$
• D. $\dfrac{2\pi }{b^{2}}\left [ a-\sqrt{\left ( a^{2}-b^{2} \right )} \right ].$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Let $f(x)=3x^{2}-7x+a, x > \dfrac{7}{6}$, the value of a such that $f(x)$ touches its inverse $f^{-1}(x)$ is
• A. $3$
• B. $-3$
• C. $\dfrac{49}{12}$
• D. $\dfrac{16}{3}$

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.