Maths

59. Let y = $\frac{e^x}{sinx}$

$\therefore \frac{dy}{dx}=\frac{sinx\frac{d{e}^x}{dx}-e^x\frac{d}{dx}sinx}{si{n}^{2}x}$

$=\frac{sinx{e}^x-e^{x}cosx}{si{n}^{2}x}$

$=e^x\frac{(sinx-cosx)}{si{n}^{-2}x.}$

58. Kindly go through the solution

57. Kindly go through the solution

56. Kindly go through the solution

12. Let z = -i

Then $\overline{z}$ = i

And |z|² = (-1)² = 1

So, multiplicative inverse of –i is given by

z^-1 = $\frac{\overline{z}}{\left|z\right|}$ = $\frac{i}{1}$ = I = 0 + i1

11. Let Z = 4 – 3i

Then $\overline{z}$ = 4 + 3i

And, z|² = 4²+ (-3)²= 16 + 9 = 25

Hence, z^-1 = $\frac{\overline{z}}{{\left|z\right|}^2}$ = $\frac{4+3i}{25}$ = $\frac{4}{25}$ + $i\frac{3}{25}$

55. Kindly go through the solution

10.${\left(-2-\frac{1}{3}i\right)}^3$

= ${\left[-\left(2+\frac{1}{3i}\right)\right]}^3$

= ${\left(-1\right)}^3\left[{\left(2+\frac{1}{3}i\right)}^3\right]$

= $(-8)\left[8+\frac{i^3}{27}+3.2.\frac{1}{3}i\left(2+\frac{i}{3}\right)\right]$ [since, (a + b)³ = a³ + b³ + 3ab(a + b)]

= (–1) $\left[8-\frac{i}{27}+(2i*2)+\left(2i*\frac{i}{3}\right)\right]$ [since, i³ = i².i = –i and i² = –1]

= (–1) $\left[8-\frac{i}{27}+4i+\frac{2i^2}{3}\right]$

= (–1) $\left[8-\frac{2}{3}+i\left(4-\frac{1}{27}\right)\right]$

= $-\frac{22}{3}$ $–\frac{i107}{27}$

54. Kindly go through the solution