Tyrell Singleton

2023-02-22

The value of the expression $\left({\mathrm{tan}}^{4}x+2{\mathrm{tan}}^{2}x+1\right){\mathrm{cos}}^{2}x$ , when $x=\frac{\pi }{12}$ is equal to
$A\right)4\left(2-\sqrt{3}\right);\phantom{\rule{0ex}{0ex}}B\right)4\left(\sqrt{2}+1\right);\phantom{\rule{0ex}{0ex}}C\right)16{\mathrm{cos}}^{2}\frac{\pi }{12};\phantom{\rule{0ex}{0ex}}D\right)16{\mathrm{sin}}^{2}\frac{\pi }{12}$

Makenna Martinez

The appropriate choices are
A) $4\left(2-\sqrt{3}\right)$
D $16{\mathrm{sin}}^{2}\frac{\pi }{12}$
$\left({\mathrm{tan}}^{4}x+2{\mathrm{tan}}^{2}x+1\right){\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}=\frac{\left(1+{\mathrm{tan}}^{2}x{\right)}^{2}}{{\mathrm{sec}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+{\mathrm{tan}}^{2}x{\right)}^{2}}{1+{\mathrm{tan}}^{2}x}\phantom{\rule{0ex}{0ex}}=1+ta{n}^{2}x$
Putting $x=\frac{\pi }{12}$
$=1+ta{n}^{2}\frac{\pi }{12}\phantom{\rule{0ex}{0ex}}=1+\left(2-\sqrt{3}{\right)}^{2}\phantom{\rule{0ex}{0ex}}=4\left(2-\sqrt{3}\right)$
Also,
$16{\mathrm{cos}}^{2}\frac{\pi }{12}\phantom{\rule{0ex}{0ex}}=4\left(\frac{\sqrt{3}+1}{\sqrt{2}}{\right)}^{2}\phantom{\rule{0ex}{0ex}}=4\left(2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}16{\mathrm{sin}}^{2}\frac{\pi }{12}=4\left(\frac{\sqrt{3}-1}{\sqrt{2}}{\right)}^{2}\phantom{\rule{0ex}{0ex}}=4\left(2-\sqrt{3}\right)$

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