( +i )- ( -i ) is equal to [RPET 2000] - Vidyadip

MCQ

$\cosh (\alpha +i\beta )-\cosh (\alpha -i\beta )$ is equal to [RPET 2000]

A
$2\,\,\sinh \,\alpha \,\,\sinh \,\beta $
B
$2\,\,\cosh \,\alpha \,\,\cosh \,\beta $
✓
$2i\,\,\sinh \,\alpha \,\,\sin \,\beta $
D
$2\,\,\cosh \,\alpha \,\,\cos \,\beta $

✓

Answer

Correct option: C.

$2i\,\,\sinh \,\alpha \,\,\sin \,\beta $

C

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Complex Numbers→Complex Numbers — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

The mean of $5$ observations is $5$ and their variance is $124$ . If three of the observations are $1, 2$ and $6$; then the mean deviation from the mean of the data is

A bag contains $6$ red, $5$ white and $4$ black balls. Two balls are drawn. The probability that none of them is red, is

The two points $A$ and $B$ in a plane are such that for all points $P$ lies on circle satisfied $\frac{{PA}}{{PB}} = k$, then $k$ will not be equal to

If $\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\text{x}\Big)=\lambda\sec2\text{x},$ then:

In a rectangle $A B C D$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $A B C D$ has equation $2 x-y+4=0$. Then, the area of the rectangle is

Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime}(a)=2$ and $f(a)=4$. Then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ equals ...... .

The value of $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} $is

There are $'m\ '$ copies each of $'n\ '$ different books in a university library. The number of ways in which one or more than one book can be selected is:

The solution of $tan\,\, 2\theta\,\, tan\theta = 1$ is

The expression $\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}+\ldots+\frac{(2011)^2+1}{(2011)^2-1}$ lies in the interval