Question

Find the value of $\theta (1+\sec 20)(1+\sec 40)(1+\sec 80)$, when $\theta =\frac{\pi }{32}$,

Answer 1

$\Theta \left[\right(1+\sec {20}^o\left)\right(1+\sec {40}^o\left)\right(1+\sec {80}^o\left)\right]$

$=\Theta \left[\right(11+\cos {20}^o\left)\right](1+\cos {40}^o)(1+\cos {80}^o)\left(\frac{1}{\cos {40}^o\cos {80}^o\cos {20}^o}\right)]$

$=\Theta \left[\right(2{\cos }^{2}10\left)\right(2{\cos }^{2}20\left)\right(2{\cos }^{2}40\left)\frac{1}{\cos {40}^o\cos {80}^o\cos {20}^o}\right]$

$\Theta \left[\frac{8{\cos }^{2}10{\cos }^{2}20{\cos }^{2}40}{\cos 40\cos 80\cos 20}\right]$

$=\Theta \left[8\cos 10\cos 20\cos 40\left(\frac{\cos {10}^o}{\cos {80}^o}\right)\right](\cos {80}^o=\sin {10}^o)$

$=\Theta \left[8\cos 10\cos 20\cos 40\cot 10\right]$

$=\Theta \left[8\cos 10\cos 20\cos \right(60-20\left)\cos \right(60+20\left)\frac{\cot 10}{\cos 80}\right]$

$=\Theta \left[8\cos {10}^o\frac{1}{4}\cos {60}^o\frac{\cot {10}^o}{\cos 80}\right]$

$=\Theta \left[\frac{{10}^o}{\cos {80}^o}\cot {10}^o\right]=\frac{\cos 10}{\sin 10}\cot 10={\cot }^{2}10$

$\because \cot \left(10\right)=5.67128182$

${\cot }^3{10}^o=32.160$

$\therefore \Theta {\cot }^2{10}^o[also,\Theta =\frac{\pi }{32}]$

$\frac{\pi }{32}\left(32.160\right)$

$\Rightarrow \frac{80}{32}\times 32.160=\frac{5788.8}{32}$

$=180.9$