Question #abcc5

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Question #abcc5

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Answer 1 · 2016-06-21T07:40:22

52. $(1 - {cos}^{2} θ) (1 + {cot}^{2} θ)$ $(1-cos^2theta)(1+cot^2theta)$

$= {sin}^{2} θ \cdot {csc}^{2} θ$ $=sin^2theta*csc^2theta$
$= 1$ $=1$

53. $cos (2 x - 3 y)$ $cos(2x-3y)$
$= cos 2 x cos 3 y + sin 2 x sin 3 y$ $=cos2xcos3y+sin2xsin3y$

54. $sin 2 x cos x - cos 2 x sin x$ $sin2xcosx-cos2xsinx$

$sin (2 x - x) = sin x$ $sin(2x-x)=sinx$

55. $\frac{1 - cos 2 x}{sin 2 x}$ $(1-cos2x)/(sin2x)$

$= \frac{2 {sin}^{2} x}{2 sin x cos x}$ $=(2sin^2x)/(2sinxcosx)$

$= \frac{sin x}{cos x} = tan x$ $=sinx/cosx=tanx$

56. $cos 4 x$ $cos4x$
$= 2 {cos}^{2} 2 x - 1$ $=2cos^2 2x-1$
$= 2 {(2 {cos}^{2} x - 1)}^{2} - 1$ $=2(2cos^2x-1)^2-1$
$= 4 {cos}^{4} x - 8 {cos}^{2} x + 2 - 1$ $=4cos^4x-8cos^2x+2-1$
$:.cos4x=4cos^4x-8cos^2x+1..(1)$

Given

$cos4x=pcos^4x+qcos^x+r...(2)$

Comparing (1) and (2)

we have
$p=4,q=-8,r=1$

57.
$costheta/(1-sintheta)$

$=(cos^2(theta/2)-sin^2(theta/2))/(cos^2(theta/2)+sin^2(theta/2)-2sin(theta/2)cos(theta/2))$

$=(cos^2(theta/2)-sin^2(theta/2))/(cos(theta/2)-sin(theta/2))^2$

$=(cos(theta/2)+sin(theta/2))/(cos(theta/2)-sin(theta/2))$

Dividing numerator and denominator by $cos(theta/2)$
we get
$=(1+tan(theta/2))/(1-tan(theta/2))$

$=(t+1)/(1-t)$

Putting $tan(theta/2)=t$

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