# Question #abcc5

Jun 21, 2016

52.$\left(1 - {\cos}^{2} \theta\right) \left(1 + {\cot}^{2} \theta\right)$

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

53.$\cos \left(2 x - 3 y\right)$
$= \cos 2 x \cos 3 y + \sin 2 x \sin 3 y$

54.$\sin 2 x \cos x - \cos 2 x \sin x$

$\sin \left(2 x - x\right) = \sin x$

55.$\frac{1 - \cos 2 x}{\sin 2 x}$

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

$= \sin \frac{x}{\cos} x = \tan x$

56.$\cos 4 x$
$= 2 {\cos}^{2} 2 x - 1$
$= 2 {\left(2 {\cos}^{2} x - 1\right)}^{2} - 1$
$= 4 {\cos}^{4} x - 8 {\cos}^{2} x + 2 - 1$
$\therefore \cos 4 x = 4 {\cos}^{4} x - 8 {\cos}^{2} x + 1. . \left(1\right)$

Given

$\cos 4 x = p {\cos}^{4} x + q {\cos}^{x} + r \ldots \left(2\right)$

Comparing (1) and (2)

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

57.
$\cos \frac{\theta}{1 - \sin \theta}$

$= \frac{{\cos}^{2} \left(\frac{\theta}{2}\right) - {\sin}^{2} \left(\frac{\theta}{2}\right)}{{\cos}^{2} \left(\frac{\theta}{2}\right) + {\sin}^{2} \left(\frac{\theta}{2}\right) - 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}$

$= \frac{{\cos}^{2} \left(\frac{\theta}{2}\right) - {\sin}^{2} \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)} ^ 2$

$= \frac{\cos \left(\frac{\theta}{2}\right) + \sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right) - \sin \left(\frac{\theta}{2}\right)}$

Dividing numerator and denominator by $\cos \left(\frac{\theta}{2}\right)$
we get
$= \frac{1 + \tan \left(\frac{\theta}{2}\right)}{1 - \tan \left(\frac{\theta}{2}\right)}$

$= \frac{t + 1}{1 - t}$

Putting $\tan \left(\frac{\theta}{2}\right) = t$