Maths

7. $\left[\left(\frac{1}{3}+i\frac{7}{3}\right)+\left(4+i\frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$

= $\frac{1}{3}$ + $i\frac{7}{3}$ + 4 + $i\frac{1}{3}$ + $\frac{4}{3}$ – i

= $\left(\frac{1}{3}+4+\frac{4}{3}\right)+i\left(\frac{7}{3}+\frac{1}{3}-1\right)$

= $\frac{1+12+4}{3}$ + $i\frac{7+1-3}{3}$

= $\frac{17}{3}$ + $i\frac{5}{3}$

6. $\left(\frac{1}{5}+i\frac{2}{5}\right)$ – $\left(4+i\frac{5}{2}\right)$

= $\frac{1}{5}$ + $i\frac{2}{5}$ – 4 – $i\frac{5}{2}$

= $\frac{1}{5}$ – 4 + i $\left(\frac{2}{5}-\frac{5}{2}\right)$

= $\frac{1-20}{5}$ + i $\left(\frac{4-25}{10}\right)$

= $\frac{-19}{5}$ + i $\left(\frac{-21}{10}\right)$

= $\frac{-19}{5}$ – i $\frac{21}{10}$

5. (1 – i) – (–1 + i6)

= 1 – I + 1 – i6

= 2 – 7i

4. 3 (7 + i7) + I (7 + i7)

= 21 + 21i + 7i + 7i²

= 21 + 28i + 7 (–1) [since i² = –1]

= 21 + 28i – 7

= 14 + 28i

1. (5i) $\left(\frac{-3}{5}i\right)$

= 5 $\left(-\frac{3}{5}\right)$ *i²          [since i² = –1]

= –3 (–1)

= 3

So, (5i) $\left(-\frac{3}{5}i\right)$ = 3 + i0

61. Kindly go through the solution

60. Kindly go through the solution

59. We have, tan x= $\frac{-4}{3}$ , x in IInd quadrant.

Since, $\frac{\pi }{2}<x<\pi$

$\frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$

sin $\frac{x}{2}$ , cos $\frac{x}{2}$ , tan $\frac{x}{2}$ are all positive.

Now, sec²x = 1 + tan²x = 1 + ${\left(\frac{-4}{3}\right)}^2$ = 1 + $\frac{16}{9}$ = $\frac{9+16}{9}$ = $\frac{25}{9}$

secx = ±$\frac{5}{3}$

cosx = ±$\frac{3}{5}$ .

cosx = $\frac{-3}{5}$ as x is in IInd quadrant.

Now, 2 sin². = 1 cosx.    [cos 2x = 1 2 sin²x.]

2 sin² $\frac{x}{2}$ = 1 $\left(\frac{-3}{5}\right)$

2 sin² $\frac{x}{2}$ =  1 +$\frac{3}{5}$= $\frac{5+3}{5}$ = $\frac{8}{5}$ .

sin² $\frac{x}{2}$ = $\frac{8}{2*5}$ = $\frac{4}{5.}$

58. L.H.S = sin 3x + sin 2x - sin x

= sin 3x - sin x + sin 2x.

= 2 cos $\frac{3x+x}{2}$ . sin $\frac{3x-x}{2}$ + sin 2x $\left[?sinA-sinB=2cos\frac{A+B}{2}sin\frac{A-B}{2}\right]$

= 2 cos $\frac{4x}{2}$ sin $\frac{2x}{2}$ + sin 2x.

= 2 cos 2x sin x + 2 sin xcosx        [ $?$ sin 2x = 2sin xcosx]

= 2 sin x [cos 2x + cosx]

= 2 sin x $\left[2cos\frac{2x+x}{2}cos\frac{2x-x}{2}\right]$ $\left[?cosA+cosB=2cos\frac{A+B}{2}cos\frac{A-B}{2}\right]$

= 2 sin x $\left[2.cos\frac{3x}{2}cos\frac{x}{2}\right]$

= 4 sin xcos $\frac{x}{2}$ . cos $\frac{3x}{2}$ = R.H.S.