Class 11th

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.

57. L.H.S. = $\frac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}.$

Using sin A + sin B = 2 sin $\frac{A+B}{2}$ cos $\frac{A-B}{2}$

cos A + cos B = 2 cos $\frac{A+B}{2}$ cos $\frac{A-B}{2}.$

56. L.H.S = sin x + sin 3x + sin 5x + sin 7x.

= (sin x + sin 7x) + (sin 3x + sin 5x)

Using,

sin A + sin B = 2 sin $\frac{A+B}{2}$ cos $\frac{A-B}{2}.$

L.H.S. = 2. Sin $\frac{x+7x}{2}$ cos $\frac{x-7x}{2}$ + 2 sin $\frac{3x+5x}{2}$ cos $\frac{3x-5x}{2}$

= 2 sin $\frac{8x}{2}$ cos $\left(-\frac{6x}{2}\right)$ + 2 sin $\frac{8x}{2}$ cos $\left(-\frac{2x}{2}\right)$

= 2 sin 4x cos 3x + 2 sin 4x cosx.[ $?$ cos (-x) = cosx]

= 2 sin 4x[cos 3x + cosx]

Using cos A + cos B = 2 cos $\frac{\mathrm{A\; +}B}{2}$ cos $\frac{A-B}{2}.$

So, L.H.S. = 2 sin 4x $\left[2·cos\frac{3x+x}{2}cos\frac{3x-x}{2}\right].$

= 2 sin 4x $\left[2·cos\frac{4x}{2}·cos\frac{2x}{2}\right]$

= 4 sin 4x. cos 2x cosx = R.H.S.

55. L.H.S = (cos x-cos y)² + (sin x- sin y)²

= ${\left[-2sin\frac{x+y}{2}sin\frac{x-y}{2}\right]}^2+{\left[2cos\frac{x+y}{2}sin\frac{x-y}{2}\right]}^2$

= 4 sin² $\left(\frac{x+y}{2}\right)$ sin² $\left(\frac{x-y}{2}\right)$ + 4 cos² $\left(\frac{x+y}{2}\right)$ sin² $\left(\frac{x-y}{2}\right)$

= 4 sin² $\left(\frac{x-y}{2}\right)$ $\left[si{n}^2\left(\frac{x+y}{2}\right)+co{s}^2\left(\frac{x+y}{2}\right)\right]$

= 4 $si{n}^2\left(\frac{x-y}{2}\right)$ . [ $?$ sin²∅ + cos²∅= 1]

= R.H.S.

54. L.H.S. = (cos x + cos y)² + (sin x- sin y)²

Using,

cos A + cos B = 2 cos $\frac{A+B}{2}$ cos $\frac{A-B}{2}$

sin A - sin B = 2 cos $\frac{A+B}{2}$ sin $\frac{A-B}{2.}$

L.H.S. = ${\left[2\cdot cos\frac{x+y}{2}cos\frac{x-y}{2}\right]}^2+{\left[2cos\frac{x+y}{2}sin\frac{x-y}{2}\right]}^2$ .

= 4. cos² $\left(\frac{x+y}{2}\right)$ cos² $\left(\frac{x-y}{2}\right)$ . + 4 cos² $\left(\frac{x+y}{2}\right)$ sin² $\left(\frac{x-y}{2}\right)$ .

= 4 cos²$\left(\frac{x+y}{2}\right)$ $\left[co{s}^2\left(\frac{x-y}{2}\right)+si{n}^2\left(\frac{x-y}{2}\right)\right]$

= 4 cos²$\left(\frac{x+y}{2}\right)$ [ $?$ cos²θ+ sin²qθ  = 1].

= R.H.S.