Ncert Solutions Maths class 11th

41. L.H.S. = cos 4x.

= cos 2 (2x)

= 1 – 2 Sin² (2x) [ cos 2x = 1 – 2 Sin²x]

= 1 – 2 [2 sin xcosx]² [ sin 2x = 2 sin xcos x]

= 1 – 2 [4 sin²xcos²x]

= 1 – 8 sin²xcos²x

= R.H.S.

40. $$L.H.S. = tan 4x

We know that,

$tan2=\frac{2tan}{1-ta{n}^2}.$ , we can write

$L.H.S=tan2(2x)=\frac{2tan2x}{1-ta{n}^{2}2x.}$

$=\frac{2\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}{1-{\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}^2}$

$=\frac{\frac{4tanx}{\left(1-ta{n}^{2}x\right)}}{\frac{{\left(1-ta{n}^{2}x\right)}^2-4ta{n}^{2}x}{{\left(1-ta{n}^{2}x\right)}^2}}$

$=\frac{4tanx*\left(1-ta{n}^{2}x\right)}{1+ta{n}^{4}x-2ta{n}^{2}x-4ta{n}^{2}x}$

$=\frac{4tanx\left(1-ta{n}^{2}x\right)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

= R.H.S.

39. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x.

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – (cot 2x + cot x) [cot (2x + x)]

We know that,

$cot(A+B)=\frac{co\mathrm{tA}co\mathrm{tB}-1}{co\mathrm{tA}+co\mathrm{tB}}$ we can write

$L.H.S=cotxcot2x-(cot2x+cotx)\left[\frac{cot2x·cotx-1}{cot2x+cotx}\right]$

= cot x cot 2x – cot 2x cot x + 1

= R.H.S.

38. L.H.S$=\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x.}$

$=\frac{(cos4x+cos2x)+cos3x}{(sin4x+sin2x)+sin3x}.$

$=\frac{2cos\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)+cos3x}{2sin\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)sin3x}.$

$=\frac{2cos\frac{6x}{2}cos\frac{2x}{2}+cos3x}{2sin\frac{6x}{2}cos\frac{2x}{2}+sin3x}$

$=\frac{2cos3xcosx+cos3x}{2sin3xcosx+3x}$

$=\frac{cos3x}{sin3x}*\frac{(2cosx+1)}{(2cosx+1)}$

$=cot3x=R.H.S$

37. $=\frac{sinx-sin3x}{si{n}^{2}x-co{s}^{2}x}$

$=\frac{2cos\left(\frac{x+3x}{2}\right)sin\frac{x-3x}{2}}{-\left(co{s}^{2}x-si{n}^{2}x\right)}$

$=\frac{2cos\frac{4x}{2}sin\left(-\frac{2x}{2}\right)}{-cos2x}\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]$

$=-\frac{2cos2xsinx}{-cos2x}[?sin(-x)=-sinx]$

= 2 sin x

= R.H.S.

36. $=\frac{sinx+sin3x}{cosx+cos3x}$

$=\frac{2\left(\frac{x+3x}{2}\right)cos\left(\frac{x-3x}{2}\right)}{2cos\left(\frac{x+3x}{2}\right)cos\left(\frac{x-3x}{2}\right)}$

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

$=\frac{sin2x}{cos2x}$

$=tan2x=$R.H.S

35. L.H.S: = $L.H. S:=\frac{sinx-siny}{cosx+cosy}$

$=\frac{2cos\frac{x+y}{2}sin\frac{x-y}{2}}{2cos\frac{x+y}{2}cos\frac{x-y}{2}}$

$=\frac{sin\left(\frac{x-y}{2}\right)}{cos\left(\frac{x-y}{2}\right)}$

$=tan\left(\frac{x-y}{2}\right)=R.H.S.$ = R.H.S.

34.$=\frac{sin5x+sin3x}{cos5x+cos3x.}$

$=\frac{2sin\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}{2cos\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}$

$=\frac{sin\frac{8x}{2}cos\frac{2x}{2}}{cos\frac{8x}{2}cos\frac{2x}{2}}$

$=\frac{sin4x}{cos4x}=tan4x=$R.H.S

33.$=\frac{cos9x-cos5x}{sin17x-sin3x}$

$=\frac{-2sin\left(\frac{9x+5x}{2}\right)sin\left(\frac{9x-5x}{2}\right)}{2cos\left(\frac{17x+3x}{2}\right)sin\left(\frac{17x-3x}{2}\right)}$

$=\frac{-sin\frac{14x}{2}sin\frac{4x}{2}}{cos\frac{20x}{2}sin\frac{14x}{2}}=\frac{-sin7xsin2x}{cos10xsin7x}=\frac{-sin2x}{cos10x}$

= R.H.S

32. L.H.S$=cot4x(sin5x+sin3x)$

$=cot4x\left[2sin\frac{8x}{2}cos\frac{2x}{2}\right]$

$=\frac{cos4x}{4x}*2sin4xcosx$

$=2·cos4x·cosx$

R.H.S$=cotx[sin5x-sin3x]$

$=cotx\left[2cos\frac{8x}{2}·sin\frac{2x}{2}\right]$

$=\frac{cosx}{sinx}·2cos4xsinx$

$=2·cos4x·cosx.$

Hence, L.H.S. = R.H.S.